Question 1124380
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In order to solve this, one has to assume that two different letters cannot stand for the same digit.


Given that, we first look at A + A + A = CA, which is to say 3*A = CA.


Since the result is a two digit number, A must be larger than 3 because 3 times 3 is only 9.  Also, A must be 9 or smaller because 9 is the largest single digit.  Then looking at a multiplication table in the 3s column where the multiplier is 4 or larger, the only result where the second digit is the same as the multiplier is 3 X 5 = 15.  Therefore A is 5 and C is 1.


Since 3 times RS is also a 2 digit number with two different digits, RS must be 32 or less because it can't be 33 and anything larger would have a result of three digits, and the smallest 2 digit number is 10 so it must be at least that.  We want to find a multiple of 3 where the tens digit of the multiplier is the same as the ones digit of the product.


For multipliers with 1 as a tens digit, the only multiplier that qualifies is 17:  3 X 17 = 51.  For multipliers with 2 as a tens digit, the only multiplier that qualifies is 24:  3 X 24 = 72.  And for multipliers with 3 as a tens digit, the qualifier is 31:  3 X 31 = 93.


But we have already identified C = 1, so 17 and 31 are disqualified, so RS is 24 and PR is 72.


SCRAP = 41257
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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