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Find the digit represented by ? in the following problem: 37? × 35 = 13090. No calculators, please.
Last edited by Sophie_A (2009-10-07 01:56:16)
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Hi, Sophie_A.
My answer is ...solved by mental calculation.
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Hey! Your first answer, and then your 'modular arithmetic' answer disappeared, Bobby.
I haven't heard of 'modular arithmetic' and I was going to try to follow your explanation.
My method simply was to divide the answer (which is only short) by 35 in my head...which was quite easy because 35 is only a 2-digit number.
A longer problem would be more difficult for my method.
Last edited by phrontister (2009-10-06 13:25:30)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Hi phrontister;
Sorry, had to delete because of 2 stupid errors. First I got the wrong answer because I can't subtract 10 from 14. Then I put 8 when I meant 4.
Here is what I wanted to say.
First notice it has to be a 0,2,4,6, or 8.
13090 mod 9 = 4
Start trying 0,2,4,6,8 looking for the one that yields a 4.
370 * 35 = 1 * 8 = 8 wrong
372 * 35 = 3 * 8 = 6 wrong
374 * 35 = 5 * 8 = 4 bingo!
And an easier way. Do a 2 digit mult 37 * 35 *10 =12950 which is easy and can be done in your head and add 70 for 372 and another 70 for 374, until you get 13090.
Last edited by bobbym (2009-10-06 17:12:51)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thanks guys! My method is as follows.
35 × ? is a number ending in 0, say X0. Write down the 0 and carry the X over. Then 35 × 7 = 245, and this added to the carried-over X is a number ending in 9. Thus X must be a number ending in 4, so it must be 4, 14, or 24. Therefore 35 × ? must be 40, 140, or 240. Only one of these is divisible by 35, namely 140 = 35 × 4. Hence ? = 4.
Does this make sense?
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Hi Bobby & Sophie_A,
Bobby: Sorry, but I'm not following what you're trying to do there. My only experience with mod is with the Excel "mod" function as it helped with my NOHOW and Telephone/YOB puzzle solutions - but that's all I understand about it.
Sophie_A: I tried your method on a problem I made up (386 x 85 = 32810) but it doesn't seem to work on that one. I'm probably doing something wrong, because I don't think I've understood your method.
With your original problem another easy option is to double the answer to 26180 and divide the result by 10 and then 7 to get 374.
If calculations have to be done totally mentally (ie, no looking at the problem during calculations), I'd probably prefer the single-digit division by 7, otherwise I think that division by the two-digit 35 (as per my second post) is easier (for me). There's not much in it.
Last edited by phrontister (2009-10-07 01:02:26)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Hi phrontister;
Modulo is based on a simple idea. We say that
read 40 is congruent to 4 mod 9. Means that when I divide 40 by 9, I am left with a remainder of 4. Or that 40 is of the form 9 t + 4 for some t. Along with being very useful in number theory it also is useful for computation. Without getting fancy I will explain what I did.
First I find that:
Meaning 13090 divided by 9 leaves a remainder of 4. Easy to get for mod 9 just add all the digits up and take them mod 9. Here I get 1 + 3 + 9 = 13
Means 13 divided by 9 leaves a remainder of 4. Another nice feature of modular arithmetic is how it simplifies computations. Supposing I wanted to test whether this multiplication was correct.
Using modular arithmetic;
We can work with the mods now instead of the numbers, this has the advantage that we are always computing with small numbers.
Therefore that calculation has an error.
I tested 370 * 35, 372 * 35, 374*35 ... making use of the fact that
So I never had to work with any number bigger than 8. Even if it would have been a problem with hundreds of digits the calculations reduce to single digit numbers.
And an easier way. Do a 2 digit mult 37 * 35 *10 =12950 which is easy and can be done in your head and add 70 for 372 and another 70 for 374, until you get 13090.
I still think that the above idea is the easiest way.
Last edited by bobbym (2009-10-07 01:20:59)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Sophie_A: I tried your method on a problem I made up (386 x 85 = 32810) but it doesn't seem to work on that one. I'm probably doing something wrong, because I don't think I've understood your method.
Your new problem is: 38? × 85 = 32810.
Start with 85 × ?. This ends with a 0, say X0. Write down 0 and carry the X. Next, 85 × 8 = 680. This plus the carried-over X must end with a 1, so X must be 1, 11, 21, 31, 41, 51, 61, 71. Hence 85 × ? must be 10, 110, 210, 310, 410, 510, 610, 710. Only one of these is divisible 85; divide it by 85.
It works perfectly fine for me.
Last edited by Sophie_A (2009-10-07 02:00:16)
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Hi Sophie_A
I used modular arithmetic as above and ? = 6.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Bobby & Sophie_A,
Bobby: Thanks for that explanation! Very clear, and a clever solution method. Once I got my head around the concept I found that doing the calculations of the two problems in my head was quite easy.
And an easier way. Do a 2 digit mult 37 * 35 *10 =12950 which is easy and can be done in your head and add 70 for 372 and another 70 for 374, until you get 13090.
Yes, that's not too hard mentally, but I find it easier to deduct 12950 from 13090 to get 140, and dividing that by 35 to get 4.
Sophie_A: Ok - thanks. Got it. A few too many figures for my head to hold, though. As I see it, that method gets harder as the number of digits in the multiplier increases. eg, 1543? x 9155 = 141334890 is too hard for me to do in my head your way...but you may have some other shortcut to deal with that. Finding ? mentally by using Bobby's mod method is no harder than the other two shorter problems (apart from doing a few more digit sums).
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Hi phrontister;
Glad you got it. Knew you would.
The 37 * 35 is a snap if you use a little algebra and know some squares. For instance:
37 * 35 = (x - 1) (x + 1) with x = 35, (x - 1) (x + 1) = (x^2 - 1) = 36^2 = 1296 - 1 = 1295 Or if you don't know 36^2 = 1296 use.
(x + 7) (x + 5) = x^2 + 12 x + 35 with x = 30 = 900 + 360 +35 = 1295
Last edited by bobbym (2009-10-08 10:50:36)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Interesting, Bobby...more ways than one to skin a cat.
I'd opt for these, which only have a few figures rattling around in my head to try to keep under control:
37 × 35: 37 × 7 = 259, × 10 = 25900, ÷ 2 = 1295
38 × 85: 85 × 40 = 3400, 85 × 2 = 170, 3400 - 170 = 3230
But multiplying these 2-digit numbers together in the usual way isn't much harder.
Last edited by phrontister (2009-10-08 17:54:09)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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