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March 27, 2017

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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?

  • ARATMETICS - ,

    the nth term in the first:
    an= 2+3n
    the mth term in the second:
    yn= 3+7m

    if an=ym
    an-ym=0
    3n-7m=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: 3×3 + 7 = 16 remainder 2, and 2×3 + 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
    3n-7y=1 or 3n=477
    solve for n, m and you have it.

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