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 Post subject: stats and problem solving #46Posted: Mon Aug 23, 2010 12:22 am

Joined: Thu Jul 08, 2010 11:09 pm
Posts: 33
i factored 14 and 4 as follows

2X7 , 2X2

I get 64 and 28

which means 65 and 29 are possible.

how do you get the 84 from the factorization?

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 Post subject: Re: stats and problem solving #46Posted: Wed Sep 01, 2010 9:17 am

Joined: Thu Jul 03, 2008 2:13 pm
Posts: 117
Hey TJ,

If you break down the common multiples of 14 and 4 that way you'd recognize that you need a number divisible by 2*2*7 = 28. Any number that contains those factors will work:

28 (2*2*7)
56 (2*2*7*2)
84 (2*2*7*3)

Because they limit it to 2-digit numbers, there won't be that many possibilities, so for a question like this is may be just as easy to just write out the multiples of 14 (since there will be fewer) and add the remainder of 1 to each:

14 --> 15
28 --> 29
42 --> 43
56 --> 57
70 --> 71
84 --> 85
98 --> 99

Then check to see which of those will yield a remainder of 1 when divided by 4:

29 (7*4 + 1)
57 (14*4 +1)
85 (21*4 + 1)

So there are three numbers that meet both conditions.

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