Transcript Jacques Inaudi (1867

How many seconds in 11 years?
How many seconds in 11 years?
Ans. 346,896,000
How many seconds in 11 years?
Ans. 346,896,000
Less than 4 seconds by a 7 year old
What is the square root of
106,929?
What is the square root of
106,929?
Ans. 327
What is the square root of
106,929?
Ans. 327
4 seconds by 8 year old
Find the number whose
cube less 19
multiplied by its cube
shall be equal to the
cube of 6.
Find the number whose
cube less 19
multiplied by its cube
shall be equal to the
cube of 6.
Ans. 3
2 seconds by 13 year old
Find the number whose
cube less 19
multiplied by its cube
shall be equal to the
cube of 6.
Ans. 3
2 seconds by 13 year old
(3³-19) x 3³ = 6³
Child Prodigy
Child Prodigy

is someone who is a master of one or more
skills or arts at an early age.
Child Prodigy


is someone who is a master of one or more
skills or arts at an early age.
someone who by the age 11 displays expert
proficiency or a profound grasp of the
fundamentals in a field usually only
undertaken by adults.
Famous Math Prodigies
Galois, Euler, Gauss, Pascal
Zerah Colburn (1804-1840)



Born the fifth of seven children.
Parents were farmers in Vermont
Born with six digits on both hands and both
feet.

The supernumerary digits had been in the family
for four generations.

With very little schooling and not being able to
read or write, Zerah at the age of 6 began
repeating the multiplication tables to himself.

Zerah began performing in public exhibitions
at the age of 6.
Zerah Colburn (1804-1840)
Questions performed:
Admitting the distance between Concord and Boston to be
65 miles, how many steps must I take in going this
distance, allowing that I go three feet at a step?
Zerah Colburn (1804-1840)
Questions performed:
Admitting the distance between Concord and Boston to be
65 miles, how many steps must I take in going this
distance, allowing that I go three feet at a step?
The answer of 114,400 was given in 10
seconds.
Zerah Colburn (1804-1840)
How many days and hours since the Christian
era commenced, 1811 years (Zerah usually
assumed a 365-day year and a 30-day
month)?
Zerah Colburn (1804-1840)
How many days and hours since the Christian
era commenced, 1811 years (Zerah usually
assumed a 365-day year and a 30-day
month)?
Answered in 20 seconds
661,015 days and 15,864,360 hours.
Zerah Colburn (1804-1840)
When asked which two numbers multiplied
together produce 1242, Zerah gave answers
as fast as he could say them:
Zerah Colburn (1804-1840)
When asked which two numbers multiplied
together produce 1242, Zerah gave answers
as fast as he could say them:
54 and 23, 9 and 138, 3 and 414, 6 and 207,
27 and 46, 2 and 621.
Zerah Colburn
After his fathers passing in
December 1822, Zerah
returned home and pursued
a simple life. The rest of his
days would see him make
astronomical calculations
for observatories, engage in
ministerial duties, and teach
modern and classical
languages and literature.
On March 2, 1840, the
Reverend Zerah Colburn
died, a husband and father
of three daughters.
George Parker Bidder (1806-1878)

Born on June 14, 1806, to a
stonemason in Moreton
Hampstead, England.

When Bidder was enrolled at
the village school at the age of
six, he found that it was not
much to his taste.

Bidder began to teach himself
to count fives and tens and then
set about to learn the
multiplication table with the use
of marbles and peas.
George Parker Bidder

His early days saw Bidder spend many hours with a local
blacksmith.
 As the months passed, Bidder still had not received any formal
instruction. While working with the blacksmith, Bidder would be
given new ideas from people who would come to test his powers.
People would continually encourage Bidder to improve and
master his peculiar faculty until the time when his talent was
almost incredible.

At the age of 10, Bidder reached a point where he could multiply 12
places of figures with 12 figures.

Bidder’s father soon saw the financial promise that his son’s talent
could generate. Withdrawing him from school, Bidder’s father took
his son about the country for the purpose of exhibition.
George Parker Bidder
Here are some typical questions put to and answered by
Bidder in his exhibitions during the years 1815-1819.
If the moon be distant from the Earth 123,256 miles and
sound travels at a rate of 4 miles per minute, how long
would it be before the inhabitants of the moon could hear
of the battle of Waterloo?
George Parker Bidder
Here are some typical questions put to and answered by Bidder in
his exhibitions during the years 1815-1819.
If the moon be distant from the Earth 123,256 miles and
sound travels at a rate of 4 miles per minute, how long
would it be before the inhabitants of the moon could hear
of the battle of Waterloo?
Ans. 2 days, 9 hrs and 34 min in less than a minute.
George Parker Bidder
If the pendulum of a clock vibrates the distance of 9¾
inches in a second of time, how many inches will it
vibrate in 7 years, 14 days, 2 hours, 1 minute and 56
seconds?
George Parker Bidder
If the pendulum of a clock vibrates the distance of 9¾
inches in a second of time, how many inches will it
vibrate in 7 years, 14 days, 2 hours, 1 minute and 56
seconds?
With each year being 365 days, 5 hours, 48 minutes and
55 seconds,
the answer, in less than a minute, was 2,165,625,744 ¾
inches.
George Parker Bidder
If the globe is 24,912 miles in
circumference, and a balloon travels 3,878
feet in a minute, how long would it be in
travelling round the world?
George Parker Bidder
If the globe is 24,912 miles in
circumference, and a balloon travels 3,878
feet in a minute, how long would it be in
travelling round the world?
Ans. in 2 minutes – 23 days, 13 hours, 18 min
George Parker Bidder
Accomplishments:




Played a significant role in the construction of Norway’s first railway.
Served as engineer-in-chief of the Royal Danish railway.
Advisor for the Metropolitan Board of London regarding draining and purification of
the river Thames.
Of all his accomplishments and endeavours, Bidder is most known for his
construction and development of the Victoria Docks.
When one thinks of the early nineteenth century, during the time of
great engineering accomplishments, one must not overlook one of
the foremost engineers of the time, George Parker Bidder.
Up until the last days of his life, Bidder had
retained his calculating abilities. When in
conversation with his friend, a query was
suggested that if the speed of light was
190,000 mi/s, and the wavelength of the
red rays at 36,918 to an inch, how many of
its waves must strike the eye in one
second? As his friend takes out a pencil in
an attempt to write out the calculations,
Bidder says: “You need not work it
out…the number of vibrations will be
444,433,651,200,000”.
Two days later, on September 28, 1878,
George Parker Bidder died.
“He is probably the most outstanding mental
calculator of all peoples and all time”.
“He is probably the most outstanding mental
calculator of all peoples and all time”.
Johann Dase was the one who uttered these
words, and of the person he was referring
to…
Johann Dase (1824-1861)
“He is probably the most outstanding mental
calculator of all peoples and all time”.
Johann Dase was the one who uttered these
words, and of the person he was referring
to…
he was referring to himself.
Johann Dase (1824-1861)



Born in Hamburg, Germany on June 23,
1824, the son of a distiller.
Very little is known of Dase’s ancestry. As for
Dase himself, he began schooling at the age
of two and a half years.
Although he began at an early age, Dase
attributes his ability to later practice and not
his early instruction.
Johann Dase (1824-1861)
At the age of 15, Dase began travelling through
Germany, Denmark and England performing in
public exhibitions.
In 1840, while in Vienna, Dase was introduced to
scientific work. Under the guidance of a
mathematics professor, Dase was shown how to
compute π. Having worked on this problem for
nearly two months, Dase had successfully
computed π to 205 places.
Johann Dase (1824-1861)
What could Dase do?
Johann Dase (1824-1861)
What could Dase do?

Dase was able to count at a glance the number of peas thrown on a
table and instantly added the spots on a group of dominoes.
Johann Dase (1824-1861)
What could Dase do?

Dase was able to count at a glance the number of peas thrown on a
table and instantly added the spots on a group of dominoes.

He could multiply mentally two numbers each of twenty figures in 6
min; of forty figures in 40 min and one hundred figures in 8 hours.
Johann Dase (1824-1861)
What could Dase do?

Dase was able to count at a glance the number of peas thrown on a
table and instantly added the spots on a group of dominoes.

He could multiply mentally two numbers each of twenty figures in 6
min; of forty figures in 40 min and one hundred figures in 8 hours.

He extracted mentally the square root of a number of 100 figures in
52 minutes.
Johann Dase (1824-1861)
When the number 935,173,853,927 was given, Dase
was proficient enough to repeat it forwards and
backwards after just glancing at it for a mere second.
Dase had offered to multiply this number by any number
offered. When 7 was chosen, Dase immediately replied
6,546,216,977,489. An hour later, just before he was to
depart from his exhibition, Dase was asked if he could
still recall the number that was discussed earlier. Dase
instantaneously repeated the number forward and
backward.
Johann Dase (1824-1861)
Dase passed away in 1861 at the age of 37.
What can be said about Dase is that he desperately desired to
produce something of value for the mathematics and science
community.
In the end, all that this math prodigy wished for was to leave his
mark on the world.
Jacques Inaudi (1867-1950)




Born to a poor Italian family on October 13,
1867.
Spent most of his early youth tending to
sheep.
At the age of 6, began to calculate in an
attempt to compensate for his boredom while
attending to the livestock.
By the age of 7, Inaudi was able to multiply 5
figures by 5 figures. With such a special
talent having been discovered, Inaudi and
his elder brother travelled to many major
cities across Europe demonstrating his
abilities in public exhibitions.
Jacques Inaudi (1867-1950)

Inaudi’s exhibition program consisted of 6 questions:
Jacques Inaudi (1867-1950)

Inaudi’s exhibition program consisted of 6 questions:
1. a subtraction involving two 21-digit numbers
Jacques Inaudi (1867-1950)

Inaudi’s exhibition program consisted of 6 questions:
1. a subtraction involving two 21-digit numbers
2. the addition of five numbers of 6 digits each
Jacques Inaudi (1867-1950)

Inaudi’s exhibition program consisted of 6 questions:
1. a subtraction involving two 21-digit numbers
2. the addition of five numbers of 6 digits each
3. squaring a 4-digit number
Jacques Inaudi (1867-1950)

Inaudi’s exhibition program consisted of 6 questions:
1. a subtraction involving two 21-digit numbers
2. the addition of five numbers of 6 digits each
3. squaring a 4-digit number
4. a division (the size of the numbers are not specified)
Jacques Inaudi (1867-1950)

Inaudi’s exhibition program consisted of 6 questions:
1. a subtraction involving two 21-digit numbers
2. the addition of five numbers of 6 digits each
3. squaring a 4-digit number
4. a division (the size of the numbers are not specified)
5. the cube root of a 9-digit number
Jacques Inaudi (1867-1950)

Inaudi’s exhibition program consisted of 6 questions:
1. a subtraction involving two 21-digit numbers
2. the addition of five numbers of 6 digits each
3. squaring a 4-digit number
4. a division (the size of the numbers are not specified)
5. the cube root of a 9-digit number
6. the fifth root of a 12-digit number
Jacques Inaudi (1867-1950)

Inaudi’s exhibition program consisted of 6 questions:
1. a subtraction involving two 21-digit numbers
2. the addition of five numbers of 6 digits each
3. squaring a 4-digit number
4. a division (the size of the numbers are not specified)
5. the cube root of a 9-digit number
6. the fifth root of a 12-digit number
plus a few questions which asks him to find which day of the week a
date falls on. (e.g., October 5, 1888 = Friday)
Jacques Inaudi (1867-1950)

When asked to find a two
digit number such that the
difference between 4
times the first digit and
three times the second is
seven and that when
reversed is the number
reduced by 18…
Jacques Inaudi (1867-1950)

When asked to find a two
digit number such that the
difference between 4
times the first digit and
three times the second is
seven and that when
reversed is the number
reduced by 18…
Inaudi, in two minutes,
replied that there were no
solutions.
Jacques Inaudi (1867-1950)

When asked to find a two
digit number such that the
difference between 4
times the first digit and
three times the second is
seven and that when
reversed is the number
reduced by 18…
Inaudi, in two minutes,
replied that there were no
solutions.

When asked to find the
number whose square
and cube roots differ by
18, his answer of 729
was given in one minute.
Jacques Inaudi (1867-1950)
One of Inaudi’s remarkable traits was his tremendous
memory for figures. In a venture to try and memorize a
100-figure number, Inaudi learned the first 36 numbers in
90 seconds, the first 57 numbers in 4 minutes, 75 figures
in 6 minutes and the whole number in 12 minutes. Within
a day or two after the venture, he was able recall the
number in its entirety.
Jacques Inaudi (1867-1950)
Although Inaudi did not attend any educational institutions nor
did he reach great heights like George Parker Bidder,
Jacques Inaudi did have a long and successful career as a
performer appearing in the United States and all over Europe.
In 1950, while in relative poverty, Inaudi died at the age of 83.
Inaudi will always be remembered as the mild, modest, calm
and very reserved man who, until the final days of his life,
continued to amuse his neighbours with his powers.
How long do these powers last?
Do they stay with the individual until their dying
day or do they merely fade as the years pass?
How long do these powers last?
Do they stay with the individual until their dying
day or do they merely fade as the years pass?

Like for any language, when one fails to speak it, the ability to recall the
language may be that much more difficult.
How long do these powers last?
Do they stay with the individual until their dying
day or do they merely fade as the years pass?

Like for any language, when one fails to speak it, the ability to recall the
language may be that much more difficult.

For those prodigies who excelled later in life, much can be said about their
ability to prevent any intrusion of other interests interfere with their everyday
lives.
How long do these powers last?
Do they stay with the individual until their dying
day or do they merely fade as the years pass?

Like for any language, when one fails to speak it, the ability to recall the
language may be that much more difficult.

For those prodigies who excelled later in life, much can be said about their
ability to prevent any intrusion of other interests interfere with their everyday
lives.

This focus allows the individual to develop their gift.
How long do these powers last?
Do they stay with the individual until their dying
day or do they merely fade as the years pass?

Like for any language, when one fails to speak it, the ability to recall the
language may be that much more difficult.

For those prodigies who excelled later in life, much can be said about their
ability to prevent any intrusion of other interests interfere with their everyday
lives.


This focus allows the individual to develop their gift.
If the prodigy wishes to, the unlimited amount of time available can really
allow the prodigy to hone in on their abilities.
How long do these powers last?
Do they stay with the individual until their dying
day or do they merely fade as the years pass?

Like for any language, when one fails to speak it, the ability to recall the
language may be that much more difficult.

For those prodigies who excelled later in life, much can be said about their
ability to prevent any intrusion of other interests interfere with their everyday
lives.




This focus allows the individual to develop their gift.
If the prodigy wishes to, the unlimited amount of time available can really
allow the prodigy to hone in on their abilities.
Mental arithmetic requires:
- no instruments, or apparatus
- no audible practice that might disturb members of the family
- no information, much can be found by asking or absorbing
The young individual may, at any time of the day, carry on with their
practice and research.
How long do these powers last?
Do they stay with the individual until their dying
day or do they merely fade as the years pass?

When calculating ability falls off, it may well be due to
loss of motivation and hence failure to keep in practice.
How long do these powers last?
Do they stay with the individual until their dying
day or do they merely fade as the years pass?

When calculating ability falls off, it may well be due to
loss of motivation and hence failure to keep in practice.

For those whose gift disappeared later in life, they had
no recollection of the methods used in childhood to
obtain the once easy solutions to the once trivial
questions.
How do you account for lightning quick
calculation?
How do you account for lightning quick
calculation?

Gauss states that in order to make lightning quick
calculations, you will need:
How do you account for lightning quick
calculation?

Gauss states that in order to make lightning quick
calculations, you will need:
1.
a powerful memory
How do you account for lightning quick
calculation?

Gauss states that in order to make lightning quick
calculations, you will need:
1.
a powerful memory
How do you account for lightning quick
calculation?

Gauss states that in order to make lightning quick
calculations, you will need:
1.
a powerful memory
2.
a real ability for calculation
How do you account for lightning quick
calculation?

Gauss states that in order to make lightning quick
calculations, you will need:
1.
a powerful memory
2.
a real ability for calculation
How do you account for lightning quick
calculation?

Gauss states that in order to make lightning quick
calculations, you will need:
1.
a powerful memory
2.
a real ability for calculation
If, for any reason, the arithmetical prodigy loses interest in
calculation, or the opportunity to practice it, his power is likely to
diminish or disappear entirely. In this respect, mental calculation is
like piano-playing, or any other highly specialized activity dependent
on long practice.
How do you account for lightning quick
calculation?
The more shortcuts…the better.
How do you account for lightning quick
calculation?
The more shortcuts…the better.
Types of shortcuts
1. Arithmetical
How do you account for lightning quick
calculation?
The more shortcuts…the better.
Types of shortcuts
1. Arithmetical
2. Psychological
Why are there far fewer recorded
cases of female prodigies?
Why are there far fewer recorded
cases of female prodigies?

Apparent reasons for this discrepancy can be attributed to
underreporting, lack of opportunity to develop the talent, and lack of
encouragement for continued advancement.
Why are there far fewer recorded
cases of female prodigies?

Apparent reasons for this discrepancy can be attributed to
underreporting, lack of opportunity to develop the talent, and lack
encouragement for continued advancement.

For many parents, the thought of taking their daughters across the
country to make professional appearances was somewhat
unattractive. This can be attributed to the fact that very few women
gained distinction, particularly in fields associated with mathematics
and calculation.
 most females having such talents would most likely have been
ignored or more importantly been discouraged from displaying
it.
Why are there far fewer recorded
cases of female prodigies?

Unlike the young men who would spend their young days with
blacksmiths or tending to the family flock, very few girls had the
opportunity to isolate themselves where they could then develop any
calculating ability.

They were not sent out on lonely vigils with sheep; their chores were
apt to be carried out in the presence of others.

Too much human interaction is detrimental to the development of
mental calculation.
Top Five Women Prodigies/
Mathematicians in History
Top Five Women Prodigies/
Mathematicians in History
5.
Mary Fairfax Somerville (1780-1872)
- Scottish and British - Mathematician
- known as the "Queen of Nineteenth Century Science," she
fought family opposition to her study of math, and not only
produced her own writings on theoretical and mathematical
science, she produced the first geography text in England.
Top Five Women Prodigies/
Mathematicians in History
4.
Sophie Germain (1776-1830)
- French - mathematician
-
She studied geometry to escape boredom during the French
Revolution when she was confined to her family's home, and
went on to do important work in mathematics, especially her
work on Fermat's Last Theorem.
Top Five Women Prodigies/
Mathematicians in History
3. Maria Gaetana Agnesi (1718-1799)
- Italian (Milan) - mathematician
-
Oldest of 21 children and a child prodigy who studied
languages and math, she wrote a textbook to explain math to
her brothers which became a noted textbook on mathematics.
She was the first woman appointed a university professor of
mathematics.
Top Five Women Prodigies/
Mathematicians in History
2.
Elena Cornaro Piscopia (1646-1684)
- Italian (Venice) - mathematician, theologian
- She was a child prodigy who studied many languages,
composed music, sang and played many instruments, and
learned philosophy, mathematics and theology. Her
doctorate, a first, was from the University of Padua, where
she studied theology. She became a lecturer there in
mathematics.
Top Five Women Prodigies/
Mathematicians in History
1.
Hypatia of Alexandria
(355 or 370–415)
- Greek - philosopher, astronomer, mathematician -
She was the salaried head of the Neoplatonic
School in Alexandria, Egypt, from the year 400.
Her students were pagan and Christian young men
from around the empire.
How do they do that?
How do they do that?

Day-Date Calculations
One problem favoured by many calculating prodigies is to name the
day of the week for any given date.
It is necessary to memorize a special value or code for each month
of the year, thus:
How do they do that?

Day-Date Calculations
One problem favoured by many calculating prodigies is to name the
day of the week for any given date.
It is necessary to memorize a special value or code for each month
of the year, thus:
Month
Value
Month
Value
January
3
July
2
February
6
August
5
March
6
September
1
April
2
October
3
May
4
November
6
June
0
December
1
How do they do that?

Day-Date Calculations
One problem favoured by many calculating prodigies is to name the
day of the week for any given date.
It is necessary to memorize a special value or code for each month
of the year, thus:
Month
Value
Month
Value
January
3
July
2
February
6
August
5
March
6
September
1
April
2
October
3
May
4
November
6
June
0
December
1
In leap years, values for January and February are reduced by one.
How do they do that?
Also, Day Values are as follows:
Day
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Value
1
2
3
4
5
6
0
What day does October 5, 1888 fall on?
What day does October 5, 1888 fall on?
1.
Take the last 2 digits of the year – 88
What day does October 5, 1888 fall on?
1.
2.
Take the last 2 digits of the year – 88
Add a quarter (one fourth of 88 is 22)
which makes the total 110.
What day does October 5, 1888 fall on?
Month
Value
June
0
September, December
1
April, July
2
January, October
3
May
4
August
5
Feb, March, November 6
1.
2.
3.
Take the last 2 digits of the year – 88
Add a quarter (one fourth of 88 is 22)
which makes the total 110.
Add the index for the month (October = 3)
which brings our total to 113
What day does October 5, 1888 fall on?
Month
Value
June
0
September, December
1
April, July
2
January, October
3
May
4
August
5
Feb, March, November 6
1.
2.
3.
4.
Take the last 2 digits of the year – 88
Add a quarter (one fourth of 88 is 22)
which makes the total 110.
Add the index for the month (October = 3)
which brings our total to 113
Add the day of the month (5) and we now
have 118.
What day does October 5, 1888 fall on?
Month
Value
June
0
September, December
1
April, July
2
January, October
3
May
4
August
5
Feb, March, November 6
1.
2.
3.
4.
Day
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Value
1
2
3
4
5
6
0
5.
Take the last 2 digits of the year – 88
Add a quarter (one fourth of 88 is 22)
which makes the total 110.
Add the index for the month (October = 3)
which brings our total to 113
Add the day of the month (5) and we now
have 118.
Divide this number by 7 and we get a
remainder of 6.
Based on the index for the day, a
reminder of 6 represents Friday.
This calculation works for dates concerned with
the 19th century (1801-1900).
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 46²?
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 46²?
1.
Take the difference between 25 and 46
46 – 25 = 21
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 46²?
1.
Take the difference between 25 and 46
46 – 25 = 21
The number 21 (in hundreds) is the first 2 numbers of the solution.
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 46²?
1.
Take the difference between 25 and 46
46 – 25 = 21
The number 21 (in hundreds) is the first 2 numbers of the solution.
2.
Take the difference between 50 and 46
50 – 46 = 4
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 46²?
1.
Take the difference between 25 and 46
46 – 25 = 21
The number 21 (in hundreds) is the first 2 numbers of the solution.
2.
Take the difference between 50 and 46
50 – 46 = 4
Now, square that difference and we have 4² = 16.
The number 16 gives the last 2 numbers of the solution.
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 46²?
1.
Take the difference between 25 and 46
46 – 25 = 21
The number 21 (in hundreds) is the first 2 numbers of the solution.
2.
Take the difference between 50 and 46
50 – 46 = 4
Now, square that difference and we have 4² = 16.
The number 16 gives the last 2 numbers of the solution.
Therefore, 46² = 2116 or 2100 + 16 = 2116.
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 29²?
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 29²?
1.
29 – 25 = 4
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 29²?
1.
29 – 25 = 4
2.
50 – 29 = 21
21² = 441
Squaring Numbers

For any numbers from 25 to 50…
e.g., What is 29²?
1.
29 – 25 = 4
2.
50 – 29 = 21
21² = 441
Therefore,
400 + 441 = 841
Squaring Numbers

For any numbers from 50 to 100…
e.g., What is 88²?
Squaring Numbers

For any numbers from 50 to 100…
e.g., What is 88²?
1.
Take the difference between 88 and 50
88 – 50 = 38
Squaring Numbers

For any numbers from 50 to 100…
e.g., What is 88²?
1.
Take the difference between 88 and 50
88 – 50 = 38
Then, doubling 38 will give us 76
Squaring Numbers

For any numbers from 50 to 100…
e.g., What is 88²?
1.
Take the difference between 88 and 50
88 – 50 = 38
Then, doubling 38 will give us 76
2.
Take the difference between 100 and 88
100 – 88 = 12
Squaring Numbers

For any numbers from 50 to 100…
e.g., What is 88²?
1.
2.
Take the difference between 88 and 50
88 – 50 = 38
Then, doubling 38 will give us 76
Take the difference between 100 and 88
100 – 88 = 12
Now, square that difference and we have 12² = 144.
Squaring Numbers

For any numbers from 50 to 100…
e.g., What is 88²?
1.
Take the difference between 88 and 50
88 – 50 = 38
Then, doubling 38 will give us 76
2.
Take the difference between 100 and 88
100 – 88 = 12
Now, square that difference and we have 12² = 144.
Therefore, 7600 + 144 = 7744 or 88² = 7744
Cubing Numbers
To find the cube of any 2-digit number, it would be
advantageous to memorize the cubes of the numbers from
1 to 9.
If one wishes to find the value of 62³…
1.
2.
3.
Put down the cube of the first term in thousands.
6³ = 216 therefore 216 000
Put down the cube of the last term
2³ = 8
thus, adding 216 000 and 8 will produce 216 008.
Add to this the product of 62 and 36 (36 = 6 x 2 x 3)
62 x 6 x 2 x 3 = 2232
Therefore,
216 008
+22 32
238 328
62³ = 238 328
Cubing Numbers
Find the value of 93³.
Cubing Numbers
Find the value of 93³.
1.
9³ = 729
Cubing Numbers
Find the value of 93³.
1.
9³ = 729
2.
3³ = 27
Cubing Numbers
Find the value of 93³.
1.
9³ = 729
2.
3³ = 27
Therefore, 729 000 + 27 = 729 027
Cubing Numbers
Find the value of 93³.
1.
9³ = 729
2.
3³ = 27
Therefore, 729 000 + 27 = 729 027
Now, 93 x 9 x 3 x 3 = 7533
Cubing Numbers
Find the value of 93³.
1.
9³ = 729
2.
3³ = 27
Therefore, 729 000 + 27 = 729 027
Now, 93 x 9 x 3 x 3 = 7533
As a result,
729 027
+ 75 33
804 357
Therefore, 93³ = 804 357.
Gifted Children Speak Out
Gifted Children Speak Out
What do you think being gifted means?
Gifted Children Speak Out
What do you think being gifted means?
It means you can do lots of things without help from grownups.
Girl, 10
I think smart and gifted are totally different. Being smart is just being able to
answer questions and answer dates. Being gifted means you gave an
imagination and spirit and you are able to think creatively.
Girl, 10
Gifted means being selected to attend a resource room because of your
behaviour and your ability to think and learn a lot easier than others.
Boy, 9
I think being gifted means having a special gift from God, I feel that if you
are gifted, you are on Earth to fulfill a need that other people can’t fulfill.
Girl, 12
Gifted Children Speak Out
What is your reaction to being called gifted?
Gifted Children Speak Out
What is your reaction to being called gifted?
I don’t mind being called gifted as long as I’m not stereotyped as being
perfect.
Boy, 9
I think the word gifted is perfect, because it means we have a ‘gift’ to
understand things others don’t.
Girl, 13
I don’t like being called gifted; it’s embarrassing and it’s like bragging.
Boy, 9
Sometimes people think ‘gifted’ means stuck up and they think that you are
going to make fun of their grades because they don’t make as good grades
as you do.
Girl, 13
Gifted Children Speak Out
Are you gifted?
Gifted Children Speak Out
Are you gifted?
I do think I am smarter than most kids my age, but in only one way: I put my
brain to use and make it do what everyone’s brain can do if they would try to
do it, or care.
Girl, 9
I’ve never really considered myself a genius but yes, I think I’m smart
because I always seem to know the answer to the questions no matter what
the questions might be.
Boy, 13
I really don’t think I am any more gifted than any of my friends; I just work
very hard at everything I do and usually I do very well.
Girl, 12
I don’t think that I am gifted because I can always learn something from
others.
Boy, 10