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
My approach was almost the same but I got lost in the end.
... This must be div by 10.
From here I just picked up letter C cuz it was a four, obviously, i was wrong (I was running out of time).