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17 Feb 2021, 07:30
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A
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A 3 digit palindrome is a 3 digit number (not starting with zero) which reads the same backwards as forwards. For example 171. The sum of all even 3 digit palindromes, is
(A) 22000
(B) 22380
(C) 22400
(D) 25700
(E) 26700
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 22000
(B) 22380
(C) 22400
(D) 25700
(E) 26700
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
17 Feb 2021, 09:16
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Bunuel
To make a 3 digit number, we need to fill these three places _ _ _
For number to be palindrome, the unit digit must be same as Hundreds place
Units or Hundred's place can be filled by {2, 4, 6, 8} i.e. 4 ways
Tens place can be filled in 10 ways using any digit from 0 to 9
Total even three digit palindromes = 4*10*1 = 40
Now, Sum of 10 palindromes with unit/hundreds plac as 2 will be 100*10*2+10*(0+1+2+3---+9) + 10*2
Now, Sum of 10 palindromes with unit/hundreds plac as 4 will be 100*10*4+10*(0+1+2+3---+9) + 10*4
Now, Sum of 10 palindromes with unit/hundreds plac as 6 will be 100*10*6+10*(0+1+2+3---+9) + 10*6
Now, Sum of 10 palindromes with unit/hundreds plac as 8 will be 100*10*8+10*(0+1+2+3---+9) + 10*8
Sum of all tehse 40 numbers = 100*10*(2+4+6+8) + 10*4*(45) + 10*(2+4+6+8) = 1000*20+1800+200 = 22000
Answer: Option A
07 Mar 2021, 07:21
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Bunuel
Solution:
We see that the last digit of the number can be 2, 4, 6, or 8 since it has to be even, and since it’s a palindrome, the last and first digits have to match and the first digit can’t be 0. In other words, the first digit can also be 2, 4, 6, or 8. However, since the middle digit has no restriction, it can be any of the 10 digits. Therefore, there are 4 x 10 x 1 = 40 such numbers (notice that the last factor is 1 because the last digit has to match whatever the first digit is). So the question is what is the sum of these 40 numbers?
Of these 40 numbers, 10 numbers begin with the digit 2, next 10 with 4, another 10 with 6 and the remaining 10 with 8. Let’s look at the 10 numbers that begin with 2. Their sum is:
202 + 212 + 222 + … + 292 = (200 x 10) + (10 + 20 + … + 90) + (2 x 10)
Similarly, the 10 numbers that begin with 4 have a sum of:
404 + 414 + 424 + … + 494 = (400 x 10) + (10 + 20 + … + 90) + (4 x 10)
The 10 numbers that begin with 6 have a sum of:
606 + 616 + 626 + … + 696 = (600 x 10) + (10 + 20 + … + 90) + (6 x 10)
And finally, the 10 numbers that begin with 8 have a sum of:
808 + 818 + 828 + … + 898 = (800 x 10) + (10 + 20 + … + 90) + (8 x 10)
Adding all the expressions on the right hand side of the equal sign and simplifying it, we have:
(200 + 400 + 600 + 800) x 10 + (10 + 20 + … + 90) x 4 + (2 + 4 + 6 + 8) x 10
= 2000 x 10 + 450 x 4 + 20 x 10
= 20,000 + 1,800 + 200
= 22,000
Answer: A
