Let x be the least number which, when divided by 16,21,24 and 28. The reminder in each case is 5 and x is divisible by 19, what is the sum of digits of X?
1) 20
2) 5
3) 8
4) 17
Answers
Given :- Let x be the least number which, when divided by 16,21,24 and 28. The reminder in each case is 5 and x is divisible by 19, what is the sum of digits of X?
1) 20
2) 5
3) 8
4) 17
Solution :-
Prime factors of 16, 21 , 24 and 28 is ,
→ 16 = 2⁴
→ 21 = 3 * 7
→ 24 = 2³ * 3
→ 28 = 2² * 7
so,
→ LCM = 2⁴ * 3 * 7 = 336 .
now, given that, it gives remainder as 5 and divisible by 19 .
then,
→ (336q + 5) / 19 = Remainder 0 .
putting q values we get,
- q = 1 => (336*1 + 5)/19 = 341 / 19 ≠ Remainder 0 .
- q = 2 => (336*2 + 5)/19 = 677 / 19 ≠ Remainder 0 .
- q = 3 => (336*3 + 5)/19 = 1013 / 19 ≠ Remainder 0 .
- q = 4 => (336*4 + 5)/19 = 1349 / 19 = Remainder 0 .
hence,
→ sum of digits = 1 + 3 + 4 + 9 = 17 (Option 4) (Ans.)
Learn more :-
वह छोटी से छोटी संख्या बताईये जिसमे 7,9,11 से भाग देने पर 1,2,3 शेष बचे
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Given : x when divided by 16,21,24 and 28. The reminder in each case is 5 and x is divisible by 19
To find : least value of x
sum of digits of X
1) 20
2) 5
3) 8
4) 17
Solution:
x = 16A + 5
x = 21B + 5
x = 24C + 5
x = 28D + 5
=> x - 5 is atleast LCM of 16 , 21 , 24 and 28
LCM of 16 , 21 , 24 and 28
= 2 * 2 * 2 * 2 * 3 * 7
= 336
x - 5 = 336E
=> x = 336E + 5
336E + 5 is divisible by 19
Hence 336E + 5 = 19F
=> 323E + 13E + 5 = 19F
323 = 19 * 17
=> 13E + 5 = 19 ( F - 17E)
13E + 5 = 19k
=> 13E + 5 = 13k + 6k
=> 6k - 5 = 13 (E - k)
=> k = 3
13E + 5 = 19k
k = 3
=> E = 4
336E + 5 = 336(4) + 5 = 1349
Sum of digits = 1 + 3 + 4 + 9 = 17
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