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What least number must be subtracted from 1936 so that the remainder

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Bunuel
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What least number must be subtracted from 1936 so that the remainder [#permalink] New post   25 Nov 2021, 09:45
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What least number must be subtracted from 1936 so that the remainder when divided by 9, 10, 15 in each case is 7 ?

(A) 39
(B) 44
(C) 51
(D) 129
(E) 141
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IanStewart
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Re: What least number must be subtracted from 1936 so that the remainder [#permalink] New post   25 Nov 2021, 11:02
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There are other ways to do these kinds of problems when the answer choices aren't so convenient, but here we know, after subtracting, that our remainder must be 7 when we divide by 10, so the units digit will need to become 7, and we must be subtracting something ending in '9'. That means either 39 or 129 is correct, but those two numbers differ by 90, a multiple of 9 and of 15, so we'll get the same remainders by 9 and by 15 when we subtract either of them from 1936, and since we want to subtract the smallest possible number, 39 must be the answer.
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Re: What least number must be subtracted from 1936 so that the remainder [#permalink] New post   27 Mar 2024, 04:47
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧



What least number must be subtracted from 1936 so that the remainder when divided by 9, 10, 15 in each case is 7 ?

1936 - a number should give us 7 as remainder when divided by 10

Concept: Remainder by 10 of any number = units' digit of that number

=> Units' digit of 1936 - a number = 7 => we are looking for a unit's digit of 9 in the option choices
=> We can eliminate B, C, E

Let's take 39 as we are trying to find the least number
1936 - 39 = 1987

Division by 9: Remainder of a number when divided by 9 = Remainder of sum of the digits by 9
=> Remainder of 1 + 9 + 8 + 7 = Remaidner of 25 by 9 = 7 => TRUE 

Division by 10: Unit's digit = 7 => Remainder of 1987 by 10 = 7 => TRUE

Division by 15: Let's divided directly by 15, we will find that 1890 is dividible by 15
=> Remainder of 1987 by 15 = 7 => TRUE 

So, Answer will be A
Hope it helps!

Watch the following video to MASTER Remainders
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Re: What least number must be subtracted from 1936 so that the remainder [#permalink] New post   27 Mar 2024, 05:32
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Bunuel wrote:
What least number must be subtracted from 1936 so that the remainder when divided by 9, 10, 15 in each case is 7 ?

(A) 39
(B) 44
(C) 51
(D) 129
(E) 141
 

To get the general case of different divisors but common remainder, all we need to do is

\(n = LCM (9, 10, 15) * a + 7 = 90a + 7\)

So a number such that remainder upon division by 9, 10 or 15 will be 7 in each case will be of the form: 7 more than a multiple of 90, so it will definitely end in 7. 
We will get such a number after every 90 numbers i.e. 97, 187, 277 etc. Hence the smallest number that we need to subtract from 1936 will not be more than 90. Hence options (D) and (E) are out. 

To get a 7 in the units digit, we must subtract  a number ending in 9 from 1936. So the only possible option is 39.

Answer (A)

If you want to verify: 

If we were to subtract 39 (option (A)) from 1936, we would get 1897. Is this of the form 90a + 7
1897 = 1890 + 7 = 90*21 + 7
Hence 39 is the smallest number which should be subtracted from 1936 to obtain the desired result. 

Check the following video on Division and Remainders to understand this logic:https://youtu.be/A5abKfUBFSc


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Re: What least number must be subtracted from 1936 so that the remainder [#permalink]
27 Mar 2024, 05:32
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