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The sum of four consecutive two-digit odd numbers, when divided by 10,
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Bunuel
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The sum of four consecutive two-digit odd numbers, when divided by 10, [#permalink] New post  25 Jan 2021, 03:45
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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?

A. 26
B. 41
C. 53
D. 57
E. 67
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Re: The sum of four consecutive two-digit odd numbers, when divided by 10, [#permalink] New post  25 Jan 2021, 04:33
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Bunuel wrote:
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?

A. 26
B. 41
C. 53
D. 57
E. 67



Odd number may have only {1, 3, 5, 7, 9} as unit digits

4 of these digits when added, must result in a number which is a multiple of 10

i.e the 4 digits must be {1, 3, 7, 9}

so the numbers must be such as {17, 19, 21, 23} or {27, 29, 31, 33} etc

Now checking options

A. 26 EVEN so OUT
B. 41 i.e. Numbers could be 37, 39, 41, 43 which sum up to 160 which when divided by 10 leaves a perfect square 16 CORRECT!!!
C. 53
D. 57
E. 67

ANswer: Option B
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Re: The sum of four consecutive two-digit odd numbers, when divided by 10, [#permalink] New post  25 Jan 2021, 18:21
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x, x+2, x+4, x+6 numbers must take the form

4x+12 = 10*x^2

2x + 6 = 5*x^2

x can only take the value up to 4 or RHS will exceed 2 digits, which was a condition of the problem

x=4 works and 37 once you solve.

41 would be in the sequence, B
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Re: The sum of four consecutive two-digit odd numbers, when divided by 10, [#permalink]
25 Jan 2021, 18:21
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