Latest Maths Olympiad Aptitude Question SOLUTION: Q. Sum of all triple digit numbers divisible by 7.

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(#M40030614) MATHS OLYMPIAD Numerical Ability question Logic Keep an EYE

Q. Sum of all triple digit numbers divisible by 7.

Asked In Maths Olympiad Harsh Poddar (10 years ago)

Solved RAKESH Read Solution (5)

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RAKESH  (10 years ago)

The first three digit no. divisible by 7 is 105 & last is 994
so numbers are 105,112,119,...,994
sum=105+112+119+...+994
series is AP with first a=105,d=7,l=994
994=105+(n-1)*7
or 889=(n-1)*7
or 127=n-1
or n=128
so sum=(n/2)*(a+l)
=(128/2)*(105+994)
=70336

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Mahesh  (10 years ago)

The first one is 105, which can be written as 7*15
The last one is 994, which can be written as 7*142

So, a total of 128 numbers
The average is (105 + 994)/2 = 1099/2
So, total is 128 *1099/2 = 64*1099 = 74336

Answer is 74336

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Biren Kumar Das  (10 years ago)

The first one is 105, which can be written as 7*15
The last one is 994, which can be written as 7*142
So from 105 to 994, 128 numbers are present.

Ans is [(1st number+Last numbber)*Total number]/2
=[(105+994)*128]/2=74336

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SASMIEL KARADO  (10 years ago)

all three digits no. divisible by 7 are 105,112,119......994
the first no. 105=7*15
the last no. 994=7*142

the total no. of three digit numbers divisible by 7 are ((142-15)
+1)=128....(1 is added because both 12 and 142 are also included)

now putting the formula for AP
sum=n/2(a+l)
=128/2(105+994)
=70336

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AMIT JAISWAL  (10 years ago)

105,114.....998
common diff=7

994=a+(n-1)d
994=105+(n-1)7
889+7 = 7n
n=128

sum of these terms=(n/2)*(a+l)
=64*110=70336

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