Alteration of Digits in Integers Problem

"Which integers have the following property? If the final digit is deleted, the integer is divisible by the new number?

Any integer with a 0 on the end would work.

I deleted all the previous stuff.

I believe it will only work when the last digit is 0.
Do you want to see some logic behind it?

Seems like your answer 'when the last digit is 5' always works, though.

I definitely understand that any integer with the last digit is 0 will work...But any logic as to the process of which you deduced it would be much appreciated.

Can you give me an example of where you think the 5 will work?

Also Jedithious could you tell me your age or give me an idea of what level you are at.
I want to pitch my answers at an appropriate level.
Are you comfortable with algebra?
Melody

105, but I just realized that 5 actually doesn't work because the number 25 contradicts it. (25 is not divisible by 2.)

105?
10 doesn't go into 105

Definitely comfortable with algebra. Comfortable with all of Grade 11 and some of 12.

Oh dear, yes. Sorry, this nasty little problem is making my brain fuzzy. And one that does actually work is 55 obviously.

I think your answer 'any integer ending with zero' is probably indeed the only answer which doesn't have any counterexamples. You don't need to explain the logic of it...I'm so tired of this problem haha...Anyways, thank you for taking the time to help me with this.

That blows my proof. I'd better think again.

Uh oh. I wonder if there is a strict systematic method or concept that exists to solve this...Or if they just expect us to test out different theories until we find one without contradictions.

(an example of what I mean by 'concept' is 'cyclicity', which is necessary to solve problems like what is the units digit of 9 to the power of 56).

ok that makes it easy for me.
Sometimes i will try to pitch a little lower because I think some of the other forum users also like to look and understand.

Think of any large number but break it into 2 pieces 10x+y where x and y are positive integers (whole numbers) and x can't be 0, y is less than 10
When you take the last digit off you are left with x

now you want (10x+y) / x to be a whole number

(10x+y) / x = 10 + y/x

so y/x has to be a whole number
y=0 will always work
If x >= 10 then y has to be 0
If x <10 then y has to be 0 or a multiple of x
So
If x = 9 y=0 or 9 90 or 99
If x = 8 y = 0 or 8 80 or 88
If x=7 y=0or 7 70 or 77
If x=6 y=0 or 6 60 or 66
If x=5 y=0 or 5 50 or 55
If x=4 y=0 or 4 or 8 40, 44, 48
If x=3 y=0,3,6,9 30,33,36,39
If x=2 y=0,2,4,6,8 20,22,24,26,28
If x=1 y can equal any single digit number

so we have
All double or more digit numbers that end in 0 plus
99,88,77,66,55,44,48,33,36,39,22,24,26, 28,19,18,17,16,15,14,13,12 and 11

How's that
That is all of them!

(an example of what I mean by 'concept' is 'cyclicity', which is necessary to solve problems like what is the units digit of 9 to the power of 56).

Got you. The last digit is 8

That is exactly what I wanted. A way to reduce and organize this seemingly complicated idea/requirement into something as clear and workable as an algebra formula. Due to your explanation of the logic I am now fully satisfied and convinced. Thanks for taking the time to solve this and explain it.

Numbers ending in zero are the obvious case, now let´s solve the other cases
Let the variable n represent the new number you get after cutting off the last digit.
Let the variable d represent the last digit.
The original number is (10*n)+d
the problem says the original number is divisible by n
which means (10*n)+d is divisible by n
((10*n)+d)/n = 10 + d/n
that means d must be a multiple of n
for example, try the number 24
4 is a multiple of 2, so it works. Check it
cut off 4, the new number is 2, which divides into 24
11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 24, 26, 28, 33, 36, 39, 44, 48, 55, 66, 77, 88, 99 all work
and of course any number ending in 0
so it looks like we are limited to two-digit numbers and numbers ending in 0