"All positive two digit numbers ending with either 0 or 1 or 5 or 6 have their squares and cubes ending with 0/1/5/6. So there are 36 positive two-digit numbers in total that have this property."
Alan, Germany
Thursday, November 2, 2023
"My guess would be 36. There are 4 possibilities in every ten and 9 possible tens (assuming we are regarding zero as not a digit) E.g. 10, 100,1000 As well as two digit numbers ending 1, 5 and 6. 11, 121, 1331 / 15, 225, 3375 / 16, 356, 4096."
Ann,
Friday, November 3, 2023
"
(1) We only need to think about the units digit of the 2-digit number.
This gives us 10 digits to consider.
(2) We can calculate d x d and d x
d x d for each of these digits d.
0 x 0 = 0 and 0 x 0 x 0 = 0 yes 1
x 1 = 1 and 1 x 1 x 1 = 1 yes 2 x 2 = 4 and 2 x 2 x 2 = 8 no 3 x 3 = 9
and 3 x 3 x 3 = 27 no 4 x 4 = 16 and 4 x 4 x 4 = 64 no 5 x 5 = 25 and 5 x
5 x 5 = 125 yes 6 x 6 = 36 and 6 x 6 x 6 = 216 yes 7 x 7 = 49 and 7 x 7 x
7 = 343 no 8 x 8 = 64 and 8 x 8 x 8 = 512 no 9 x 9 = 81 and 9 x 9 x 9 =
729 no
Note: There is a more efficient method of calculating the units
digit of any cubed number. We can simply take the units digit of the
squared number (ignore the tens) and continue. If we start with 4, then 4
x 4 is 16. Ignore the ten, 6 x 4 = 24 so 4³ ends in a 4.
Only two
digit numbers ending in 0, 1, 5 and 6 will have squares and cubes ending in the
same digit. There are four possible ending digits and each gives 9 solutions,
giving us a total of 36 solutions.
You could also say that the two digit
number itself, it’s square and it’s cube all end in the same digit.
Anyone interested in exploring further might like to investigate ‘cyclicity’.
I can recommend this lovely video from Learners’ Planet: How to Find the Unit
Digit of Large Powers | Unit Digit Tricks | Learners' Planet
https://www.youtube.com/watch?v=ujifcj0iiEI
"Initially, I solved the puzzle the old-fashioned way, using my brain. One has to only focus on the last digit, which quickly finds numbers ending in 0, 1, 5, and 6, since when squared, the last digit is the same as the original number, and hence, will be the same when cubed. Then, to make sure these were the only choices, I did the multiplications on the other six ending digits, finding none that satisfied the criteria. Therefore, there are four numbers in each decade (for example, 10, 11, 15, 16, 20, 21, 25, 26, and so on). Finally, I wrote a Python script to verify. In Python, indexing an array with minus 1 gets the last array element.
for TwoDigit in range(10, 99, 1):
TwoDigitSquared = TwoDigit * TwoDigit
TwoDigitCubed = TwoDigitSquared * TwoDigit
if str(TwoDigitSquared)[-1] == str(TwoDigitCubed)[-1]:
print (TwoDigit, TwoDigitSquared, TwoDigitCubed)"
Iniha J Rajan, Hilversum, Netherlands
Sunday, November 5, 2023
"The answer to the puzzle of the month for November is 36. I found the pattern by first squaring and cubing the first seven 2 digit numbers. The patters is that every multiple of five's square and cube number end in the same digit. Then the number after every multiple of five's cube and square also ends in the same digit. I wrote all the multiple of five's that are 2 digits, and the number after that, then counted them. The final number was 36."
Mala, Papatoetoe High School
Wednesday, November 1, 2023
"All positive two digit numbers ending with either 0 or 1 or 5 or 6 have their squares and cubes ending with 0/1/5/6. So there are 36 positive two-digit numbers in total that have this property."
Alan, Germany
Thursday, November 2, 2023
"My guess would be 36.
There are 4 possibilities in every ten and 9 possible tens (assuming we are regarding zero as not a digit)
E.g. 10, 100,1000
As well as two digit numbers ending 1, 5 and 6.
11, 121, 1331 / 15, 225, 3375 / 16, 356, 4096."
Ann,
Friday, November 3, 2023
"
(1) We only need to think about the units digit of the 2-digit number. This gives us 10 digits to consider.
(2) We can calculate d x d and d x d x d for each of these digits d.
0 x 0 = 0 and 0 x 0 x 0 = 0 yes
1 x 1 = 1 and 1 x 1 x 1 = 1 yes
2 x 2 = 4 and 2 x 2 x 2 = 8 no
3 x 3 = 9 and 3 x 3 x 3 = 27 no
4 x 4 = 16 and 4 x 4 x 4 = 64 no
5 x 5 = 25 and 5 x 5 x 5 = 125 yes
6 x 6 = 36 and 6 x 6 x 6 = 216 yes
7 x 7 = 49 and 7 x 7 x 7 = 343 no
8 x 8 = 64 and 8 x 8 x 8 = 512 no
9 x 9 = 81 and 9 x 9 x 9 = 729 no
Note: There is a more efficient method of calculating the units digit of any cubed number. We can simply take the units digit of the squared number (ignore the tens) and continue. If we start with 4, then 4 x 4 is 16. Ignore the ten, 6 x 4 = 24 so 4³ ends in a 4.
Only two digit numbers ending in 0, 1, 5 and 6 will have squares and cubes ending in the same digit.
There are four possible ending digits and each gives 9 solutions, giving us a total of 36 solutions.
You could also say that the two digit number itself, it’s square and it’s cube all end in the same digit.
Anyone interested in exploring further might like to investigate ‘cyclicity’. I can recommend this lovely video from Learners’ Planet:
How to Find the Unit Digit of Large Powers | Unit Digit Tricks | Learners' Planet
https://www.youtube.com/watch?v=ujifcj0iiEI
The whole video is excellent, but you might like this table for your newsletter problem:
https://youtu.be/ujifcj0iiEI?si=5MVUFhr7qOwwVZB2&t=497
Thank you for sharing your lovely problem."
Rick,
Saturday, November 4, 2023
"Initially, I solved the puzzle the old-fashioned way, using my brain. One has to only focus on the last digit, which quickly finds numbers ending in 0, 1, 5, and 6, since when squared, the last digit is the same as the original number, and hence, will be the same when cubed. Then, to make sure these were the only choices, I did the multiplications on the other six ending digits, finding none that satisfied the criteria. Therefore, there are four numbers in each decade (for example, 10, 11, 15, 16, 20, 21, 25, 26, and so on).
Finally, I wrote a Python script to verify. In Python, indexing an array with minus 1 gets the last array element.
for TwoDigit in range(10, 99, 1):
TwoDigitSquared = TwoDigit * TwoDigit
TwoDigitCubed = TwoDigitSquared * TwoDigit
if str(TwoDigitSquared)[-1] == str(TwoDigitCubed)[-1]:
print (TwoDigit, TwoDigitSquared, TwoDigitCubed) "
Iniha J Rajan, Hilversum, Netherlands
Sunday, November 5, 2023
"The answer to the puzzle of the month for November is 36.
I found the pattern by first squaring and cubing the first seven 2 digit numbers. The patters is that every multiple of five's square and cube number end in the same digit. Then the number after every multiple of five's cube and square also ends in the same digit. I wrote all the multiple of five's that are 2 digits, and the number after that, then counted them. The final number was 36."