Question

Find the smallest positive integer whose cube ends in 888.

Answer 1

hey mate
hope you like the answer
please mark me as the brainliest

My Try: First of all its cube root should have Last digit as 22.

Now if x888x888 is a Perfect Cube Then

x888=888+1000x=8(111+125x)x888=888+1000x=8(111+125x)

which means 111+125x111+125x should be a Perfect cube.


You've already figured out the ones digit of the root is 22.

So, the number (10x+2)3=1000x3+600x2+120x+8(10x+2)3=1000x3+600x2+120x+8 ends with ...888...888.

This number modulo 1010 is obviously 88. So, you remove the 88 and divide by 1010. Now you got 100x3+60x2+12x100x3+60x2+12x. This number has to end in ...88...88.

Now, 100x3+60x2+12x≡12x(mod 10)100x3+60x2+12x≡12x(mod 10)

So, 12x12x has to end in an 88 and thus, xxeither ends in 44 or 99. So, now we've narrowed down possibilities to x=4,9,14,19,24...x=4,9,14,19,24... If you check these numbers, you find that x=19x=19 is the answer.

Thus, the smallest perfect cube to end in 888 is 192. (Remember the root is 10x+210x+2)

EDIT: If you want to do a bit more work, you can figure out the following:

60x260x2 will always end in ...60...60 for xx that ends in 44 or 99. (If it isn't immediately obvious, work at it a bit more).

Which means that, 12x12x must end in ...28...28 to get the desired ...88...88 - this narrows down to ALL numbers whose cubes end in ...888...888, which are 19,44,69,9419,44,69,94.

Here, an interesting pattern emerges.

All numbers of the form 10(19+25x)+210(19+25x)+2 (where xx is a non-negative integer) are roots of perfect cubes ending in ...888...888.

thanks for asking...
I think this is the correct answer please mark me as the brainliest