Dividing by 7

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
The following is a question from an Australian Mathematics Competition, and was to be done without using a calculator.

The two digit number ab (a and b are digits) is divisible by 7. If ba means the transpose of ab (for example, 31 is the transpose of 13), which of the following must also be divisible by 7?

(1)...5 X b + a (2)...3 X a + b (3)...ba + a

Note,..at least one of the above is correct.

Be sure to show all your working so as to substantiate your answer, and remember...no calculators....pen and paper ONLY!

Have fun,... :D
Ron.
 
SydneyRoverP6B said:
The following is a question from an Australian Mathematics Competition, and was to be done without using a calculator.

The two digit number ab (a and b are digits) is divisible by 7. If ba means the transpose of ab (for example, 31 is the transpose of 13), which of the following must also be divisible by 7?

(1)...5 X b + a (2)...3 X a + b (3)...ba + a

Note,..at least one of the above is correct.

Be sure to show all your working so as to substantiate your answer, and remember...no calculators....pen and paper ONLY!

Have fun,... :D
Ron.
Click to expand...

OK, seems a good one!

The technical correct way to denote 'ab' is (10 x a) + b

(1) (5 x b) + a, so by x 10, we get (50 x b) + (10 x a) = (10 x a) + (50 x b) = [(10 x a) + b] + (49 x b)

(49 x b) is divisible by 7 (49/7 = 7), and since we know (10 x a) + b is divisible by 7, it means that (5 x b) + a also divides into 7 with no remainder.

(2) (3 x a) + b = ([10 - 7] x a) + b = ([10 x a] + b) - (7 x a)

(7 x a) is divisible by 7 (7/7 = 1), (3 x a) + b must also divide exactly into 7.

3) (10 x b) + a + a = (10 x b) + (2 x a). so by x 5, we get (50 x b) + (10 x a) = (10 x a) + (50 x b) = [(10 x a) + b] + (49 x b)

As in (1), we get [(10 x a) + b] + (49 x b), which we know is divisible by 7.

They all seem to be divisible by 7!


(Shoving a huge wad of cash under "quattro" Richard's door, to pay for showing me how to complete this problem! :p)
 
Hello Darth,

That is the right answer... :D All three are divisible by 7.

A collaboration with Richard (quattro) too,...great mathematical minds working together to solve a problem,...excellent.. :D

Glad that you both enjoyed it, not to forget all the other forum members who gave it a shot as well.

Ron.
 
SydneyRoverP6B said:
Hello Darth,

That is the right answer... :D All three are divisible by 7.

A collaboration with Richard (quattro) too,...great mathematical minds working together to solve a problem,...excellent.. :D

Glad that you both enjoyed it, not to forget all the other forum members who gave it a shot as well.

Ron.
Click to expand...

Oh yes, he wouldn't have done it without my help LOL 8)

Come to think of it, I had nothing to do with it, :?: or did I do something then forgot about it?

:? Nope, I was down the pub at the time and I have witnesses :wink:
 
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