r/askmath • u/Old-Finance-9217 • 1d ago
Algebra Any numbers
I was having a discussion with an enemy about how Wonder Woman 1984 was a government plot to cover up 1984 to stop any ideas of that sort of thing and then he said that if you add up all the numbers it makes 2000. Says I, that’s 22. But that got me wondering is there any number that doesn’t end in 0 that if you multiply by another number you get 0s before the ending digit. Something like 12*x=102
Ĕ̈d̆̈ĭ̈t̆̈:̆̈ S̆̈ĭ̈n̆̈c̆̈ĕ̈ Ĭ̈ c̆̈l̆̈ĕ̈ă̈r̆̈l̆̈y̆̈ w̆̈ă̈s̆̈n̆̈t̆̈ c̆̈l̆̈ĕ̈ă̈r̆̈ ĕ̈n̆̈ŏ̈ŭ̈ğ̈h̆̈,̆̈ h̆̈ĕ̈r̆̈ĕ̈ ă̈r̆̈ĕ̈ s̆̈ŏ̈m̆̈ĕ̈ m̆̈ŏ̈r̆̈ĕ̈ ĕ̈x̆̈ă̈m̆̈p̆̈l̆̈ĕ̈s̆̈
2̆̈3̆̈1̆̈x̆̈(̆̈y̆̈)̆̈=̆̈2̆̈3̆̈0̆̈1̆̈
1̆̈1̆̈1̆̈1̆̈x̆̈(̆̈y̆̈)̆̈=̆̈1̆̈1̆̈1̆̈0̆̈1̆̈
T̆̈h̆̈ĕ̈ s̆̈ă̈m̆̈ĕ̈ n̆̈ŭ̈m̆̈b̆̈ĕ̈r̆̈ b̆̈ŭ̈t̆̈ ŏ̈n̆̈l̆̈y̆̈ ă̈ 0̆̈ b̆̈ĕ̈f̆̈ŏ̈r̆̈ĕ̈ t̆̈h̆̈ĕ̈ l̆̈ă̈s̆̈t̆̈ d̆̈ĭ̈ğ̈ĭ̈t̆̈
7
u/hansn 1d ago
I don't understand your question.
But that got me wondering is there any number that doesn’t end in 0 that if you multiply by another number you get 0s after the ending digit.
You give the example of 102, which has a 0 before the final digit. 51*2 gives you 102, and that seems like it meets your first criteria of not ending in zero.
But I'm guessing I'm not quite understanding your question. Since this makes no sense to me:
I was having a discussion with an enemy about how Wonder Woman 1984 was a government plot to cover up 1984 to stop any ideas of that sort of thing and then he said that if you add up all the numbers it makes 2000. Says I, that’s 22.
5
u/ExistentAndUnique 1d ago edited 1d ago
I don’t understand the first half of the question.
For the latter part, such numbers do exist: 11 * 91 = 1001. In fact, infinitely many numbers work (for the appropriate generalization)
If you split a number into its last digit and the remaining parts, you have written it in the form (10x+y) for some integers x and y. Adding another 0 before the last digit can be obtained by instead taking (100x+y), another 0 would be (1000x+y) and so on. Your goal is to identify whether any of these numbers is a multiple of 10x+y.
Theres a neat trick that I won’t get into, but it turns out that this will work for any value of 10x+y that is not divisible by 2 or 5. It can work for others as well (such as 22 * 91 = 2002), but I would need to think about it a bit more
6
3
3
u/tomic24 1d ago
sure, lots of them. e.g. 9x12=108, 9x112=1008...
3
1
u/peter-bone 21h ago
That's not of the format requested. The result must have the same digits as one of the numbers being multiplied, but with a zero at the penultimate position.
2
u/MezzoScettico 1d ago
OK, reading your edit I see what you're asking.
For instance with your example 231x = 2301. Let's represent 231 as 23*10 + 1. What you want is an x such that (23*10 + 1)x = 2301 = 23*100 + 1
The general expression is that you want (10a + b)x = 100a + b where b is a one digit number.
I'm pretty sure this doesn't have any integer solutions that meet all the conditions, but let me see if I can prove it. Notice that x is always going to be close to 10, a little less, because the number on the right (2301 in your example) is about 10 times the number on the left (231).
4
u/MezzoScettico 1d ago
Actually I found a solution. If b = 9, then we wnat
(10a + b)*9 = 100a + b
90a + 9b = 100a + b
8b = 10a
a = (8/10)b = (4/5)b
Since b is a single digit, this can only be an integer if b = 5, which means a = 4 is the only solution.
45 * 9 = 405
I don't think there are any other solutions possible.
1
1
1
12
u/donutello2000 1d ago
I like how everyone was so floored by the latter part of the post that they completely ignored the part about OP having a discussion with an enemy.