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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?

1. $21$
2. $25$
3. $41$
4. $67$
5. $73$

Let x =First odd Number

Sum of 4 consecutive two digit odd numbers =  x + x + 2 + x + 4 + x + 6  = 4x + 12

Let y= $(\frac{4x+12}{10})^{\frac{1}{2}}$

$\frac{4x+12}{10}=y^{2}$

$\frac{2x+6}{5}$=$y^{2}$

There are only 9 choice (17,27,37,47,57,67,77,87,97) are possible for the valve of x.

From the help of given options we can eliminate (27,47,57,77,87,97).

now check for remaining (17,37,67) choices.

put x=17

$\frac{40}{5}=8$

Here 8 is not a perfect square. so can elimate Option (1) and (2) from here.

now put x=67

$\frac{140}{5}=28$

Here 67 is not a perfect square. so can elimate Option (4) and (5) from here.

Now put x=37

$\frac{80}{5}=16$

Here 16 is a perfect square.

and four consecutive two digit odd numbers x =37

x + 2=39

x + 4 =41

x + 6  =43

Hence (3)41 is the correct choice.

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### 1 comment

How did you those 9 choices please inform me.

1