Question 1198924: The letters P, Q and R each stands for a different digit in the alphametric (PP)^2 + (PQ)^2 = QRQR. (Here, PP means a two-digit number, like 33.) If PP and PQ are consecutive numbers, then the value of P+Q + R is
Found 3 solutions by math_tutor2020, greenestamps, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Hint:
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81
Focus on the units digit of each result to get this sequence: 1, 4, 9, 6, 5, 6, 9, 4, 1
Notice we have symmetry.
Sort the values and toss out any duplicates to get: 1, 4, 5, 6, 9
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Another hint:
Q = P+1 since PP and PQ are consecutive numbers (eg: 64 and 65)
PP and PQ are of different parity (one is odd, the other even, in either order)
(PP)^2 and (PQ)^2 are of different parity. I'll leave the proof for the reader.
odd + even = odd
which means QRQR is odd, and R itself is odd
The possible values of R is from the set {1,3,5,7,9}
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
There are algebraic techniques that you could use to solve the problem, or to work parts of the problem. But the conditions of the problem allow only a very few possible solutions, so trying each of them to find the answer is as good a method as any.
The conditions of the problem say that the number PP is a 2-digit integer with both digits the same, an PQ is the next larger integer. Try each possibility:
11^2+12^2 -- don't even need to do the actual calculations; the sum is not 4 digits
22^2+23^2 = 484+529 = 1013
33^2+34^2 = 1089+1136 = 2225
44^2+45^2 = 1936+2025 = 3961
55^2+56^2 = 3025+3136 = 6161
There is our answer. P = 5; Q = 6; R = 1
ANSWER: P+Q+R = 12
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
The letters P, Q and R each stands for a different digit in the alphametric (PP)^2 + (PQ)^2 = QRQR.
(Here, PP means a two-digit number, like 33.) If PP and PQ are consecutive numbers, then the value of P+Q + R is
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First notice that since the sum QRQR is a 4-digit number, the numbers PP and PQ can not be too large.
Indeed, P must be less than 7; otherwise the sum (PP)^2 + (PQ)^2 will be 5-digit number.
Having it, there are not so many options for you to check: all possible oprtions are
(1) PP = 66 PQ = 67
(2) PP = 55 PQ = 56
(3) PP = 44 PQ = 45
(4) PP = 33 PQ = 34
(5) PP = 22 PQ = 23
(6) PP = 11 PQ = 12
It gives for (PP)^2 + (PQ)^2 these possible values
(1) 8845
(2) 6161
(3) 3961
(4) 2245
(5) 1013
(6) 265
Of these options, only (2) has the reqired form, with the number 6161.
So, P = 5, Q = 6, R = 1,
and P + Q + R = 5 + 6 + 1 = 12. ANSWER
Solved.
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