x is a positive integer less than 100. When x is divided by 5, the rem

Question

x is a positive integer less than 100. When x is divided by 5, the remainder is 4, and when x is divided by 23, the remainder is 7. What is the value of x?

A. 7
B. 30
C. 53
D. 76
E. 99

Archit3110 · Answer

Clearly option E..
 99= 5*19+4 and 99=23*4+7

gracie · Answer

x=5q+4
x=23p+7
→23p-5q=-3
if 5q has a 5 units digit,
then 23p must have a 2 units digit
try 4*23=92
92-5*19=-3 yes
p=4, q=19,
and n=99
E

thewarrior29 · Answer

X=5a+4
X=23b+7

5a has to be divisible by 5 and total has to be less than 100, so a must have to be less than 20. Take a=19, that will give u value of X=99.
Now check second equation. Divide 99 by 23 and u will find a remainder of 7. As (23*4=92). So 99 works.
 
So ans is 99.

LeoN88 · Answer

Take multiples of 23 & keep adding 7 to the number (<100). For every such number check if it's ending with 4 or 9.

gurmukh · Answer

Only option when divided by 5 gives remainder 4 is 99, hence it is the answer choice

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TheNightKing · Answer

Use options.

Only option E, which is 99, gives us remainder 4 when divided by 5. NO need to check anything else.

Answer E

BrushMyQuant · Answer

When x is divided by 5, the remainder is 4

Dividend = Divisor * Quotient + Remainder

=> x = 5*k + 4 (where k is an integer)
=> x = 5k + 4

when x is divided by 23, the remainder is 7

=> x = 23t + 7 (where t is an integer)
=> x = 5k + 4 = 23t + 7
=> 5k = 23t + 3
=> k =  \frac{23t + 3}{5}

Only those values of t which will give k also as an integer are the possible values 
=> t = 4 and k =  \frac{23*4 + 3}{5}  = 19
=> n = 5k + 4 = 5*19 + 4 = 99

So,  Answer will be E 
Hope it helps!

Watch the following video to MASTER Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

x is a positive integer less than 100. When x is divided by 5, the rem

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x is a positive integer less than 100. When x is divided by 5, the rem

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