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01 Sep 2018, 15:17
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C
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Difficulty:
95%
(hard)
Question Stats:
33% (03:09) correct
67%
(02:26)
wrong
based on 55
sessions
History
Date
Time
Result
Not Attempted Yet
The palindrome 939 is the sum of three lesser palindromes. If the three have different lengths and use only the same two positive integers,
which is the sum of these two integers?
A. 8
B. 9
C. 10
D. 11
E. 12
which is the sum of these two integers?
A. 8
B. 9
C. 10
D. 11
E. 12
01 Sep 2018, 18:56
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Answer: C
Eliminated choices which cannot be used in below structure to get 939 as described below:
8 - 1+7 / 2+6 / 3+5 /4+4
9 - 1+8 / 2+7 / 3+6 / 4+5
11 - 2+9 / 3+8 / 4+7 / 5+6
12 - 3+9 / 4+8 / 5+7 / 6+6
XYX
YY
X
However, 10 possible to be split into - 1+9 / 2+8 / 3+7 / 4+6 / 5+5 does have the combination of (1,9) which can fit into below structure and give us 939
919
11
9
Eliminated choices which cannot be used in below structure to get 939 as described below:
8 - 1+7 / 2+6 / 3+5 /4+4
9 - 1+8 / 2+7 / 3+6 / 4+5
11 - 2+9 / 3+8 / 4+7 / 5+6
12 - 3+9 / 4+8 / 5+7 / 6+6
XYX
YY
X
However, 10 possible to be split into - 1+9 / 2+8 / 3+7 / 4+6 / 5+5 does have the combination of (1,9) which can fit into below structure and give us 939
919
11
9
01 Sep 2018, 20:42
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gracie
firstly I take it that use only the same two positive integers, means use any one or both of same two positive integers
a logical and fast approach could be -
the SUM is 939 and the other two palindrome will sum up lesser than 100+10 that is <110
so the biggest palindrome is \(> 939-110\) or \(>829\) and therefore the hundreds digit will be 8 or 9
lets check with 8..
units digit of sum is 9, and that will be possible when two palindromes have 8 as units digit and third will become X9-2*8=X9-16=_3
so other digit will be 3
1) max possible sum = \(888+88+3 =979\neq{939}\)
2) other possibility = \(888+33+8 =929\neq{939}\)
so 8 is not the digit
Now let's check with 9..
units digit of sum is 9, and that will be possible when all three are ODD units digit or when two digits are even and the third is 9
But he biggest is 9X9, which is <939 so X<3, possible values 0, 1 and 2
a)when two palindromes have 0 as units digit and third will become 9.. but then we will not have 2-digit number
b) when two palindromes have 9 as units digit and third will become X9-2*9=X9-18=_1..
so other digit will be 1
possibility = \(919+11+9 =939=939\)
so the digits are 9 and 1, sum = 1+9=10
C
