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29 Apr 2026, 14:32
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How many positive integers x are there such that when x is divided by 1000, the remainder is x/9?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
30 Apr 2026, 21:38
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We know that Dividend=divisor*quotient+rem
So x=1000q+(x/9)
Solving the above gives you x=1125q
But we also know that rem=x/9
So rem=x/9=1125q/9=125q
Now we know that 0<=remainder<divisor
So 0<=125q<1000
So the range of possible values of q should be 1,2,3,4,5,6,7 in order to satisfy the equation above. Hence, the answer is 7.
We are not considering 0 as a possible value since the question stem specifies that we need positive integers
So x=1000q+(x/9)
Solving the above gives you x=1125q
But we also know that rem=x/9
So rem=x/9=1125q/9=125q
Now we know that 0<=remainder<divisor
So 0<=125q<1000
So the range of possible values of q should be 1,2,3,4,5,6,7 in order to satisfy the equation above. Hence, the answer is 7.
We are not considering 0 as a possible value since the question stem specifies that we need positive integers
19 May 2026, 13:43
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as per the given condition 1000>x/9 or x<9000-------i
also x=1000k+x/9 where k is a +ve integer
8x/9=1000k or x=9000 *k/8-------ii
only values of k that satisfy both i and ii are from 1 through 7
hence there are 7 values of x that satisfy the required condition
E
also x=1000k+x/9 where k is a +ve integer
8x/9=1000k or x=9000 *k/8-------ii
only values of k that satisfy both i and ii are from 1 through 7
hence there are 7 values of x that satisfy the required condition
E
