M40-20

Question

Lisa wrote down 10 consecutive integers on a blackboard. However, Bart, being mischievous, erased one of them. If the sum of the remaining nine numbers is 146, what is the value of the number that Bart erased?

A. 7
B. 9
C. 12
D. 19
E. 21

A. 7
B. 9
C. 12
D. 19
E. 21

Bunuel · Accepted Answer

Official Solution:

Lisa wrote down 10 consecutive integers on a blackboard. However, Bart, being mischievous, erased one of them. If the sum of the remaining nine numbers is 146, what is the value of the number that Bart erased?

A. 7
B. 9
C. 12
D. 19
E. 21

A. 7
B. 9
C. 12
D. 19
E. 21

Assume the smallest of the 10 consecutive integers is  x . Then their sum is:
  
 x + (x + 1) + ... + (x + 9) = 10x + (1 + 2 + ... 9) = 10x + \frac{1 + 9}{2}*9 = 10x + 45 .
  
Now, let's assume Bart erased the number  k . Then, we'd have:
   
 10x + 45 = 146 + k ;
 
 10x = 101 + k .
   
Since  x  is an integer, the left-hand side of the equation is a multiple of 10. Therefore, the right-hand side must also be a multiple of 10, which means the units digit of  k  must be 9. This narrows our options down to B (9) and D (19). However,  k  cannot be 9 because in that case  x  would be 11, and the erased number cannot be smaller than the smallest number in the set. Therefore, Bart must have erased the number 19.

Answer: D

Bunuel · Answer

If k = 9, then x = 11, so the consecutive integers are 11, 12, ..., 20. As you can see, this sequence does not contain 9.

Billyneutron · Answer

Hello, I could not understand the part where 9 is eliminated; pls explain, thanks

sindhureddyalla · Answer

I like the solution - it’s helpful.

naoliden · Answer

I like the solution - it’s helpful.

AVMachine · Answer

With that calculation we got x = 13; and last element x+9 = 22;

Sum of all the elements in the series = (13+22)*10/2 = 175

Now, if 19 got removed from the 175 we got = 175-19 = 156, not 146?

Bunuel · Answer

__________________
x = 12, not 13.

sg8595 · Answer

What a brainstormer!!!

Bixy34 · Answer

Hi  Bunuel , I approached this question in a different manner and I am unable to understand what did I do wrong. Here is what I did- 
10 consecutive integers= x-4, x-3, x-2, x-1, x, x+1, x+2, x+3, x+4, x+5 i.e 10x +5 should be the sum of the consecutive integers. 
Now, 10x + 5= 146 + x (x being the missing number here). 
Then I went straight to the answer choices and with 9, x was coming out to be an integer so I marked it without evaluting any options further. Now I see that even 19 would satisfy here. 
How do I approach this question from here?

Bunuel · Answer

You should use a different letter for the missing number. How can it be x again? Use a different variable for the missing number, like k, and then proceed the way shown in the official solution.

Bixy34 · Answer

Ahh, got it! My bad. Thank you for your prompt response!:)

AdarshSambare · Answer

Solution Approach:

To minimize calculation, assume the 10 consecutive numbers are:

n-4, n-3, ..., n, ..., n+5

If one number x is erased, then:

Sum of 10 numbers = 10n + 5

Given sum (after erasing) = 146 = (10n + 5) - x

10n = 141 + x

So, we are left with possible values x = 9 or 19.

Checking the possibilities:

* If x = 9:
 The sequence would be 9, 10, 11, ..., 18.
 Total sum =   –   
 = 171 - something
The difference will be lesser than 146
If you have time calculate sum of 9 numbers else leave FYI,
=   –   
= 171 – 45 = 126 
which is less than 146. Not possible.

* If x = 19:
 The numbers are valid, and the sum matches the condition.

Final Answer: 19 (Option D)

agrasan · Answer

I like the solution - it’s helpful.

MarcoPag · Answer

This is a great question that’s helpful for learning and I like the solution - it’s helpful.

Aboyhasnoname · Answer

My approach...The Answer cannot be A C E ..because...adding them will give us the units digit such as 3 or 8 or 7 ..means the avg wil be xx.3 or xx.8 or xx.7 ..Thats not possible in the case of consecutive integers... could be the average will be xx.5 when even number of consecutive integers..Only 9 and 19 gives us this...Now lets add 9 to 146..We get 155 Avg becaomes 15.5 means the middle two terms are 31 in total means 15 and 16 .they are 5th and 6th term..means Consecutive integers here start from 11...So 9 Cannot be the answer......We are left with 19.....

Bunuel

Lisa wrote down 10 consecutive integers on a blackboard. However, Bart, being mischievous, erased one of them. If the sum of the remaining nine numbers is 146, what is the value of the number that Bart erased?

A. 7
B. 9
C. 12
D. 19
E. 21

Adit_ · Answer

Took me a bit of time to solve(3.5mins), but I do believe this should be 805+ considering there are many 805+ ones that are much simpler, this requires a judicious use of elimination and precision. The way I found out was:
10/2(first+last)=

This   has to be a multiple of 5 as 10/2 results in 5 and now:

Using option only 9/19+146 leads to a multiple of 5. Pick 19:

146+19=165 We thus find first+last=33 OR

First is 12 and last is 21:

19 falls within the range of 12 and 21 and thus 19 fits perfectly.

_ankitchawla_ · Answer

I like the solution - it’s helpful.

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