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Math Team Questions That I Have; Already Did The Questions, Just Asking For Another Way To Solve


Lightmeow
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Hello,

 

I was doing my math team practice packet, and in the writers choose section, I had three questions on the problems.

 

Lets see, for the first question was this: How many integers (10<x<100) are increased by nine when their digits are reversed. So my answer was eight numbers; the numbers 12, 23, 34, 45, 56, 67, 78, and 89, and I know that is right. I had to do it all out by hand, until I noticed a pattern which was you take the last digit of the number, make that the first digit of the next number, and have the number in the tens place be one less than the one in the ones. It took me about five minutes to reach that conclusion. Now my question would be, is there a way to find out the answer to this question mathematically without having to do it all out like I did. Say if they gave me a different problem(I made this one up), like how many integers are decreased by 6 when their digits are reversed, how would I find that out using algebra.

 

The second question was, how many natural numbers give a remainder of two when divided into 83. My answer was 4 numbers, the numbers being 3, 9, 27, 81. Again, there was a simple pattern. it being 3x<83. Again, how would you do that mathematically, without having to d it all out. Say you had another problem, how many natural numbers give a remainder of three when divided into 275(I just picked a random number). Could you help me there?

 

The third question would be N is a 5 digit number 8A65B in which A and B are digits, and N is divisible by 24. What is the smallest number N can be. My answer was 82656, but I had to do it all out, and that took a while.

 

I am asking these questions, because when I go to a math meet, I only have about ten minutes to do these problems, and it took me twenty minutes to do them. Is there a way that I could make the solving of those questions go faster.

 

Thanks for you time,

 

Joshua

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For the first, you could let the first digit be x and the second be y, and use the equation 10x + y + 9 = 10y + x. Simplifying, you end up with x = y - 1, i.e. the second digit is one more than the first, which is the pattern you found. I'm not sure your second example (increasing by 6) is possible, since using the same reasoning (and assuming you're still talking about two-digit numbers), we end up with x = y - 2/3.

For the second problem, what you're looking for is the set of solutions to the equation 83 = nx + 2, or 81 = nx, i.e. x divides 81. The factors of 81 are 1, 3, 9, 27, and 81. Clearly 1 isn't a solution. For 275 and a remainder of 3, we have that x must divide 272. The factors of 272 are 1, 2, 4, 8, 16, 17, 34, 68, 136, and 272. A quick way to find these factors is to find the prime factorization of 272, which is 2^4 * 17, and realize that since each of the prime factors divides 272, any product consisting of the prime factors also divides 272.

For the last, since the number is divisible by 24, it must be divisible by both 8 and 3. The divisibility rule for 8 says a number is divisible by 8 if its last three digits are divisible by 8. So you have 65B, and you must find B. An immediate thing that pops into my mind is 640 is divisible by 8, so if we add two 8's to that, we arrive at 656 being divisible by 8. Therefore, B is 6. Now, the divisibility rule for 3 says a number is divisible by 3 if the sum of its digits is divisible by 3. At this point, we have 8A656, so we need to find A such that 8 + 6 + 5 + 6 + A = 25 + A is divisible by 3. We want the smallest number, so we can start with A = 0 and work our way up. Obviously 0 doesn't work, and letting A = 1 doesn't work either (the sum will be 26). But when A = 2, we have 8 + 2 + 6 + 5 + 6 = 27, which most certainly divides 3. Thus, 82656 is the smallest number of the form 8A65B that divides 24.

Edited by John
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