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gracie
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A palindrome is a number that reads the same forward and backward. The 08 Dec 2018, 18:33
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C
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31% (03:21) correct 69% (02:57) wrong based on 88 sessions

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A palindrome is a number that reads the same forward and backward. The digits of a three digit palindrome sum to a perfect square. How many such three digit palindromes are there?

A. 11
B. 12
C. 13
D. 14
E. 15
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C
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raffaeleprio
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A palindrome is a number that reads the same forward and backward. The 16 Apr 2021, 06:49
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Hi!

First of all let's notice that the smallest 3-digit palindrome number whose digits' sum is a perfect square is 121, which corresponds to 4.
At the same time, we see that the biggest one corresponds to 979 which is 25.
Hence, our problem turns into finding out the numbers whose sum will be :
4 9 16 25.

Now let's keep in mind that in order to have 2 equal digits, the number sum can be written as S= 2x + y so:
(hint: start from x=1 and then increase x up to run out S)
4= 1+1+2= 2+2+0 hence 121 and 202
9= 1+1+7= 2+2+5=3+3+3=4+4+1 hence 171,252,333,414
16= 4+4+8= 5+5+6= 6+6+4=7+7+2= 8+8+0 hence 484, 565, 646, 727, 808
25=8+8+9= 9+9+7 hence 898, 979

It naturally follows that the number of such numbers is 13, which leads to answering C.

IMO C!
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A palindrome is a number that reads the same forward and backward. The 11 Oct 2024, 14:16
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Why not 15?
For a number xyx sum is 2x+y which should be a perfect square; so
At x=1, y=0,2,7
At x=2, y=0,5
At x=3, y=3,9
At x=4, y=1,8
At x=5, y=6
At x=6, y=4
At x=7, y=2
At x=8, y=0,9
At x=9, y =7
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A palindrome is a number that reads the same forward and backward. The 11 Oct 2024, 14:29
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VasundharaS
Why not 15?
For a number xyx sum is 2x+y which should be a perfect square; so
At x=1, y=0,2,7
At x=2, y=0,5
At x=3, y=3,9
At x=4, y=1,8
At x=5, y=6
At x=6, y=4
At x=7, y=2
At x=8, y=0,9
At x=9, y =7
Show more

If x = 1 and y = 0, how is the sum a perfect square?
If x = 3 and y = 9, how is the sum a perfect square?
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A palindrome is a number that reads the same forward and backward. The 11 Oct 2024, 22:54
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Possible Perfect squares are 4,9,16,25

Fix the first and last digits in increasing order and keep plugging in the difference in the middle.

Perfect Square = 4 (2)

121
202

Perfect Square = 9 (4)

171
252
333
414

Perfect Square = 16 (5)

484
565
636
727
808

Perfect Square = 25 (2)

898
979

Total = 13, IMO C
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A palindrome is a number that reads the same forward and backward. The 11 Oct 2024, 23:37
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A palindrome is a number that reads the same forward and backward. The digits of a three digit palindrome sum to a perfect square. How many such three digit palindromes are there?

A. 11
B. 12
C. 13
D. 14
E. 15

So, we have hundreds and ones digit equal to each other and tens which takes value such that all the three sum to a perfect square.
We have ten digits from 0 to 9
Perfect squares are 1, 4, 9, 16, 25, 36 .... so on.

Thus, values that can occur at hundreds and ones place are 1 to 9
Values that can occur at tens place are 0 to 9

Lowest three digit number is 100 but not a palindrome, so lowest become 121(as digits of 101 are not 111 summing to perfect square) and highest is 999.

Putting a limit to highest would help in knowing what is highest value that a sum can take. Therefore, sum of 999 is 27 which is not a perfect square but caps the limit of perfect square to next lowest perfect square i.e. 25.

So, we have only 4, 9, 16 and 25 that are possible in the three digit numbers.
A. For number with 1 as hundreds and ones digit desired possibilities are 121, 171 (sum of 2 leaves tens digit minimum to 2 and max being [23(25-2)] which is not possible upto which only 7 is possible)
B. For number with 2 as hundreds and ones digit desired possibilities are 202 and 252 (sum of 4 leaves tens digit minimum to 0 and max being 9 [21(25-4)] which is not possible upto which only 5 is possible)
C. For number with 3 as hundreds and ones digit desired possibilities are 333 only (sum of 6 leaves tens digit minimum to 3 and max being 9 [19(25-6)] which is not possible)
D. For number with 4 as hundreds and ones digit desired possibilities are 414 and 484 (sum of 8 leaves tens digit minimum to 1 and max being 9 [17(25-8)] which is not possible upto which only 8 is possible)
E. For number with 5 as hundreds and ones digit desired possibilities are 565 only (sum of 10 leaves tens digit minimum to 6 and max being 9 [15(25-10)] which is not possible)
F. For number with 6 as hundreds and ones digit desired possibilities are 646 only (sum of 12 leaves tens digit minimum to 4 and max being 9 [13(25-12)] which is not possible)
G. For number with 7 as hundreds and ones digit desired possibilities are 727 only (sum of 14 leaves tens digit minimum to 0 and max being 9 [11(25-14)] which is not possible)
H. For number with 8 as hundreds and ones digit desired possibilities are 808 and 898 (sum of 16 leaves tens digit minimum to 0 and max 9(25-16) which is possible)
I. For number with 9 as hundreds and ones digit desired possibilities are 979 only (sum of 18 leaves tens digit minimum to 7 and max 9(25-18) which is not possible)

Hence, possible numbers are
2 + 2 + 1 + 2 + 1 +1 + 1 + 2 + 1 = 13

Answer B.
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