M31-21

Question

What is the value of  27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2 ?

A. 6298 
B. 6308 
C. 6318
D. 6328 
E. 6338

Bunuel · Accepted Answer

Official Solution:

What is the value of  27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2 ?

A. 6298 
B. 6308 
C. 6318
D. 6328 
E. 6338

We have 7 terms, with the middle term being  30^2 . We can express all other terms as  30-x :
  
 27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2= 
 
 =(30-3)^2+(30-2)^2+(30-1)^2+30^2+(30+1)^2+(30+2)^2+(30+3)^2 . 
  
Next, expand these expressions using the identity  (x-y)^2=x^2-2xy+y^2  for the first three terms and  (x+y)^2=x^2+2xy+y^2  for the last three terms. Notice that the  -2xy  and  2xy  terms will cancel out, leaving us with:
  
 (30^2+3^2)+(30^2+2^2)+(30^2+1^2)+30^2+(30^2+1^2)+(30^2+2^2)+(30^2+3^2)= 
 
 =7*30^2+2*(3^2+2^2+1^2)=6,300+28=6,328 .

Answer: D

danjbon · Answer

An alternate approach is to find the tens digit, since they are all different. You would get:

29
84
41
00
61
24
89

Which results in a tens digit of 2 and leads you to select D

z0rpia · Answer

I used the following rules to solve this problem in under 1 minute. I'm not sure if this is the right application of the rules but it gets me to the RIGHT answer. I applied the following rules for sum of consecutive integers to this problem which is sums of SQUARES OF consecutive integers.

Rule: The sum of k consecutive integers, with an ODD number of items, k, is always a multiple of the number of items, k.

Example: 4 + 5 + 6 + 7 + 8 = 30 , multiple of 5

Rule: The sum of k consecutive integers, with an EVEN number of items, k, is NEVER a multiple of the number of items, k.

Example: 4 + 5 + 6 + 7 + 8 + 9 = 39 , NOT a multiple of 6

In this question, there are 7 consecutive integers. In the answer choices, the only answer divisible by 7 is 6328. Answer Choice D

norovers · Answer

I approached by finding the units digit in each set and adding them together because the numbers all have different last two digits. Thus, 7*7 ends in 9, 8*8 ends 4....3*3 ends in 9.

Thus, 9+4+1+0+1+4+9=28
Only choice D ends with 28, thus it is our answer.

This approach is NOT correct.

RenanBragion · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Hitesh0701 · Answer

Hi Bunuel,
Is this correct approach to get the answer?

Bunuel · Answer

No, that approach is not valid. For example, if the expression were 27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 +  4 3^2, using the same method you’d still get a units-digit sum of 28, but the actual total is 7088.

adarsh2209 · Answer

I like the solution - it’s helpful.

Apoorv0194 · Answer

Can we use the formula here ?  /6 putting n = 33 and substracting putting n = 27

Bunuel · Answer

Yes, you can use that formula. But you need to subtract the sum up to n = 26, not 27, because the series starts from 27^2.

M31-21

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