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Bunuel
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Here is my approach =>
X= 243P+98 for some integer P
=> X= 81P+98 = 81Q +17 for some integer Q
hence if we subtract 17 => X=81Q which would be divisible by P
SMASH A
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Bunuel
A certain number X when divided by 243 leaves a remainder of 98. Which of the following can be subtracted from X to make X divisible by 81?

A. 17
B. 27
C. 37
D. 47
E. 57

Be careful.
341 is NOT the smallest possible (positive) value of X
X could equal 98, since 98 divided by 243 equals 0 with remainder 98

If we want X to be divisible by 81, we can just subtract 17 from 98 (to get 81)
Answer:
Related Video
NOTE: The relevant section starts at 2:57
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Bunuel
A certain number X when divided by 243 leaves a remainder of 98. Which of the following can be subtracted from X to make X divisible by 81?

A. 17
B. 27
C. 37
D. 47
E. 57

X= 243 + 98
X=243+100-2=341

81*4=324

So we need to substract 17 from 341 to be able to make it divisible by 81.

Answer A
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Bunuel
A certain number X when divided by 243 leaves a remainder of 98. Which of the following can be subtracted from X to make X divisible by 81?

A. 17
B. 27
C. 37
D. 47
E. 57

Given: A certain number X when divided by 243 leaves a remainder of 98.
Asked: Which of the following can be subtracted from X to make X divisible by 81?

243 = 81*3
X = 243k + 98 ; where k is an integer
X = 81(3k+1) + 17
X - 17 = 81(3k+1) ; Divisible by 81

IMO A
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Kinshook
Bunuel
A certain number X when divided by 243 leaves a remainder of 98. Which of the following can be subtracted from X to make X divisible by 81?

A. 17
B. 27
C. 37
D. 47
E. 57

Given: A certain number X when divided by 243 leaves a remainder of 98.
Asked: Which of the following can be subtracted from X to make X divisible by 81?

243 = 81*3
X = 243k + 98 ; where k is an integer
X = 81(3k+1) + 17
X - 17 = 81(3k+1) ; Divisible by 81

IMO A

Kinshook this is a much more clearer explanation, thank you!
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