sukhman wrote:The sum of four consecutive two-digit odd numbers when divided by 10 become a perfect square . Which of the following can be one of these four numbers .
A 21 B 25 C 41 D 67 E 73
Let n = smallest odd number in set
So, n+2 = 2nd odd number in set
n+4 = 3rd odd number in set
n+6 = 4th odd number in set
Sum of all 4 odd numbers = n + (n+2) + (n+4) + (n+6) = 4n + 12 =
4(n + 3)
So, as we can see,
the sum of ANY four consecutive odd integers MUST be divisible by 4.
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Now let's examine some perfect squares: 1, 4, 9, 16, 25, 36, ....
We know that, when we divide the sum of the four integers by 10, we get a perfect square.
So, the sum of the four integers must equal 10 or 40 or 90 or 160 or 250, or 360 or . . .
Since the sum of the four consecutive odd integers MUST be divisible by 4, we know that
the sum of the four integers must equal 40 or 160 or 360 or 640 or . . .
Now let's check some scenarios
If the sum of the four integers equals
40, then each of the four integers has a value near 10 (since
40/4 = 10).
None of the answer choices have a value near 10.
If the sum of the four integers equals
160, then each of the four integers has a value near 40 (since
160/4 = 40).
Answer choice C is close to 40 (looks promising)
If the sum of the four integers equals
360, then each of the four integers has a value near 90 (since
360/4 = 90).
None of the answer choices have a value near 90.
So, answer choice
C is the only answer choice that could work, so this must be the correct answer.
Aside: the values 37, 39,
41, 43 add to 160
Cheers,
Brent
