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05 Apr 2020, 21:41
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D
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Difficulty:
95%
(hard)
Question Stats:
25% (02:37) correct
75%
(02:30)
wrong
based on 104
sessions
History
Date
Time
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Not Attempted Yet
What is the sum of digits of smallest positive integer that can be written as the sum of 39 positive consecutive integers and as the sum of 41 positive consecutive integers?
A. 15
B. 18
C. 21
D. 24
E. 27
A. 15
B. 18
C. 21
D. 24
E. 27
05 Apr 2020, 22:21
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Sum of first 39 natural numbers =780
Sum of first 41 natural numbers
=861
LCM of 780 and 861 =671580
Therefore sum of the digits = 27
Option E is the answer
Posted from my mobile device
Sum of first 41 natural numbers
=861
LCM of 780 and 861 =671580
Therefore sum of the digits = 27
Option E is the answer
Posted from my mobile device
05 Apr 2020, 22:34
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Let us consider two series for this problem.
series 1: 39 consecutive integers
let the smallest positive integer of the series 1 be a
sum:
a+(a+1)+(a+2)+.......+(a+38)=39a+ 38(39)/2=39*a+741
Similarly, assume the smallest integer of the other series as b
b+(b+1)+.....+(b+40)=41*b+820
As per the question,
n=41*b+820=39*a+741 where n is the smallest positive integer.
n when divided by 41 and 39 leaves a remainder of 0
So, the smallest such number would be the L.C.M of 41 and 39 which is 1599.
sum of the digits 1+5+9+9=24
IMO D
series 1: 39 consecutive integers
let the smallest positive integer of the series 1 be a
sum:
a+(a+1)+(a+2)+.......+(a+38)=39a+ 38(39)/2=39*a+741
Similarly, assume the smallest integer of the other series as b
b+(b+1)+.....+(b+40)=41*b+820
As per the question,
n=41*b+820=39*a+741 where n is the smallest positive integer.
n when divided by 41 and 39 leaves a remainder of 0
So, the smallest such number would be the L.C.M of 41 and 39 which is 1599.
sum of the digits 1+5+9+9=24
IMO D
