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# The sum of 4 consecutive two-digit odd integers, when divided by 10,

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Intern
Joined: 31 May 2017
Posts: 26
Location: India
The sum of 4 consecutive two-digit odd integers, when divided by 10, [#permalink]

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30 Oct 2017, 08:08
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The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67

Can someone please provide me the solution to this question.

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Math Expert
Joined: 02 Aug 2009
Posts: 5774
The sum of 4 consecutive two-digit odd integers, when divided by 10, [#permalink]

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30 Oct 2017, 08:29
swatygi wrote:
The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67

Can someone please provide me the solution to this question.

the answer depends on properties of number...

first there are five odd digits... 1,3,5,7,9..
SUm of which 4 consecutive gives you multiple of 10.... it is 7+9+1+3=20..
so the numbers will be TWo with one lower TENS digit - X7 and X9 and two with one higher TENS digit - (X+1)1 and (X+1)3
so BASICALLY we are adding SUM of tens digit {x+x+(x+1)+(x+1)}*10
SuM of units digit is 7+9+1+3=20... so when you add this to TENS sum, we get $$(x+x+(x+1)+(x+1))*10 + 20 = 10x+10x+10(x+1)+10(x+1) +20=10x+10+10x+10+10(x+1)+10(x+1) = 10(x+1)+10(x+1)+10(x+1)+10(x+1)=40(x+1)$$
so $$\frac{40*(x+1)}{10} = 4*(x+1)$$ should be PERFECT SQUARE..
so x+1 can be 4 as 4*4 will be perfect square..
so numbers are 37,39,41,43

ans C

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The sum of 4 consecutive two-digit odd integers, when divided by 10,   [#permalink] 30 Oct 2017, 08:29
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# The sum of 4 consecutive two-digit odd integers, when divided by 10,

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