Question 833337: The cube of 99 is 970299. Consider the smallest two-digit number whose cube ends with the original two-digit number. The sum of the digits of the cube of that two-digit number is:
a) 18 b) 19 c)20 d)21 e)22
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Twelve-year old geniuses at the forum in the artofproblemsolving website may be able to provide a brilliant and concise explanation showing why choices b, c, d, and e are wrong. My explanation will not be so brilliant.
One strategy would be using a spreadsheet to have the computer calculate  and see if it ends in 00 for all numbers  between 11 and 99.
Without using a computer, I need a different strategy.
Algebra tells me that
 .
Since  , and  are 3 consecutive numbers, all I need to do is look for 3 consecutive integers between 10 and 100 whose product ends in 00.
If the product ends in 00, it must be a multiple of  .
Because the product is a multiple of  , one of the numbers must be a multiple of  , or else two of the numbers ( , and ) must be even (multiples of  ).
Also, one of the 3 numbers must be a multiple of  , because we could not have two multiples of  within the 3 consecutive integers.
So the products must include , or  , or  or  .
For the smallest , I should make one of the 3 integers.
The 3 consecutive integers including would be smallest if we make  . In that case, for  to be a multiple of , we would need (the only even integer of the three) to be a multiple of 4.
Since means  ,  is the smallest two-digit number whose cube ends with the original two-digit number.
Since  ,  is a multiple of 9, and the sum of its digits must be a multiple of 9. It could not be 19, or 20, or 21, or 22.
If it is one of the options given, it must be  .
|
|
|
