ARATMETICS
posted by JULIE .
THE ARITHMETIC PROGRESSION 2,5,8,11 AND 3,10,17,24... HAVE SOME COMMAN VALUES.WHAT IS THE LARGEST VALUE LESS THEN 500 THAT THEY HAVE IN COMMAN?

the nth term in the first:
an= 2+3n
the mth term in the second:
yn= 3+7m
if an=ym
anym=0
3n7m=1
3n=7m+1
this means above that 7m +1 is divisible by 3 3,6,9,12,15 (see that m=2 works, and n=5).
Lets look at the divisibility rule at the high end.
7m+1 divisible by three; 501, 498, 495, 492, .... Now, which of those minus one are divisible by 7?
489, 486, 483, ...
Another method is multiplication by 3. A number of the form 10x + y has the same remainder when divided by 7 as 3x + y. So get the leftmost digit of the original number, multiply by 3, add the next digit, get the remainder by 7, and continue from the beginning: multiply by 3, add the next digit, etc. For example, the number 371: 3Ã—3 + 7 = 16 remainder 2, and 2Ã—3 + 1 = 7. This method can be used to find the remainder of division by 7
So here subtract one, then test.
489, 486, 483, ...
3*4+8=20 r 6, 3*6+8=24 not divisible
485: 3*4+8=2r6, 3*6+5=23 not divisible
482: 3*4+*=2r6, 6*3+2=20 not divisible
479: in my head, 2r5, 15+9=24 not div
476: 2r5, 15+6=21 divisible...
so 7y+1=477, or 7y=476
3n7y=1 or 3n=477
solve for n, m and you have it.