The sum of four consecutive two digit odd numbers when divided by 10 becomes a perfect square Which of the

as the options are not mentioned so we have to find all the possible solutions.
any four consecutive odd digit number can be expressed as 2n-3,2n-1,2n+1,2n+3,
sum of them =8n,now dividing it by 10,we get 4*n/5

for 4n/5 to be a perfect square,n should be 5* (square of a number) and n10
so n can
20,45,

1. 37,39,41,43
2.87,89,91,93
8 years agoHelpfull: Yes(3) No(1)
The answer is 41
8 years agoHelpfull: Yes(1) No(1)
sum of the numbers must be divisible by 10
so units place must be 0 this can be achieved by taking odd no at units place as 7,9,1,3
take numbers in tens place such that it could be a perfect square
ans is 7,9,11,13
sum=40
40/10=4 (2^2) is a perfect square
8 years agoHelpfull: Yes(1) No(0)
Here the conceptual expertise will be possessed by the one who knows that maximum sum of any four consecutive two digits odd numbers can at most be 389. Now on twisting the question we have to think of a number less than 389 which when divided by 10 will result in a perfect square.

Hence the feasible sum can be 360, 250, 160 or 90. Now let's go with the options as answer should be a number near to ¼ of the sum. 5th option is easily eliminated as ¼ of none of the assumed sum above is near 73. It should be near 90, 63, 40, and 23 and checking further we get numbers 41 as the part of four odd numbers i.e. 37, 39, 41 and 43 as the numbers, that are odd and which add up to 160 thereby satisfying our conditions.

So 41 is the right answer.
8 years agoHelpfull: Yes(0) No(0)

The sum of four consecutive two digit odd numbers , when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?