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Thread: Numbers Everywhere!

20061226, 04:41 #1
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Numbers Everywhere!
 <LI>Find two different whole numbers such that neither number contains any zeros and their product yields 1000000000.
<LI>Four different whole numbers sum to one hundred twentyfive. If you increase one of these numbers by four, decrease the second by four, multiply the third by four, and divide the last by four, you will produce four equivalent numbers. What are the four original numbers that sum to 125?

20061226, 05:32 #2
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Re: Numbers Everywhere!
a. <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">512 * 1953125</span hide>
b. <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">1,5,24,96</span hide>

20061226, 07:35 #3
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Re: Numbers Everywhere!
b. oops! <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">2,2,25,100</span hide> <img src=/S/blush.gif border=0 alt=blush width=15 height=15>

20061226, 07:38 #4
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Re: Numbers Everywhere!
b. <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">3, 7, 23, 92</span hide>

20061226, 20:18 #5
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Re: Numbers Everywhere!
a) Just to give you the full set and the only one answer:
<span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">
1 1000000000
2 500000000
4 250000000
5 200000000
8 125000000
10 100000000
16 62500000
20 50000000
25 40000000
32 31250000
40 25000000
50 20000000
64 15625000
80 12500000
100 10000000
125 8000000
128 7812500
160 6250000
200 5000000
250 4000000
256 3906250
320 3125000
400 2500000
500 2000000
512 1953125
625 1600000
640 1562500
800 1250000
1000 1000000
1250 800000
1280 781250
1600 625000
2000 500000
2500 400000
2560 390625
3125 320000
3200 312500
4000 250000
5000 200000
6250 160000
6400 156250
8000 125000
10000 100000
12500 80000
12800 78125
15625 64000
16000 62500
20000 50000
25000 40000
31250 32000
32000 31250
40000 25000
50000 20000
62500 16000
64000 15625
78125 12800
80000 12500
100000 10000
125000 8000
156250 6400
160000 6250
200000 5000
250000 4000
312500 3200
320000 3125
390625 2560
400000 2500
500000 2000
625000 1600
781250 1280
800000 1250
1000000 1000
1250000 800
1562500 640
1600000 625
[b]<big>1953125 512 <THIS ONE</big>[b]
2000000 500
2500000 400
3125000 320
3906250 256
4000000 250
5000000 200
6250000 160
7812500 128
8000000 125
10000000 100
12500000 80
15625000 64
20000000 50
25000000 40
31250000 32
40000000 25
50000000 20
62500000 16
100000000 10
125000000 8
200000000 5
250000000 4
500000000 2</span hide>Jerry

20070103, 05:20 #6
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Re: Numbers Everywhere!
Several stabs were taken at the ([img]/forums/images/smilies/cool.gif[/img] puzzle and so far, all are incorrect. I'll post the correct solution in a day or two if no one solves it.

20070103, 05:36 #7
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Re: Numbers Everywhere!
Now you do have me stumped.
"Four different whole numbers sum to one hundred twentyfive. If you increase one of these numbers by four, decrease the second by four, multiply the third by four, and divide the last by four, you will produce four equivalent numbers. What are the four original numbers that sum to 125?"
1st number : 3
2nd number: 7
3rd number: 23
4th number: 92
3+7+23+92=125
1st number increased by 4: 7
2nd number decreased by 4: 3
3rd number multiplied by 4: 92
4th number divided by 4: 23
Same numbers as before, only in different order. Still add to 125.
Actually, there are more solutions: 8,12,21,84; 13,17,19,76; 18,22,17,68. I guess I don't understand the question.

20070103, 16:18 #8
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Re: Solution
I think the confusion is caused by the word "equivalent". "Equal" would have been more appropriate.

20070103, 16:46 #9
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Re: Solution
I see your point but the term equivalent is correct in the algebraic sense as it means "two or more expressions that have the same value"... just a comment
Jerry

20070103, 18:04 #10
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Re: Solution
Based on your directions shouldn't this be:
<span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">16 + 24 + 5 + 80 = 125.</span hide>
The numbers in yours (since you created the algebraic solution not based on the "order") is in the wrong order:
"Four different whole numbers sum to one hundred twentyfive. If you increase one of these numbers by four, decrease the second by four, multiply the third by four, and divide the last by four, you will produce four equivalent numbers. What are the four original numbers that sum to 125"
Now, when the first is increased by 4, the second is decreased by 4, the third is multiplied by 4 and the fourth is divided by 4 all give you <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">20</span hide>. In your order this does not work out... <img src=/S/smile.gif border=0 alt=smile width=15 height=15>
Steve

20070104, 00:40 #11
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Re: Solution
OK

20070105, 15:29 #12
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Solution
Perhaps the wording of the puzzle is a bit confusing. I apologize. The one and only correct solution along with the math behind it is noted below...
<span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">The objective is to find four numbers that add to 125 whereby those four numbers become equivalent (the same number) after adding 4 to one, subtracting 4 from another, multiplying one by 4 and dividing the last by 4. If this is more clear than the original instructions, stop here before proceeding to the final solution...</span hide>
<span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">The math behind the solution: (x+4) + (x4) + (4x) + (x/4) = 125
6x + x/4 = 125
25x/4 = 125
x=20</span hide>
<span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">In the end, you have: 16 + 24 + 5 + 80 = 125. You can see now, that by adding 4 to one of the numbers, subtracting 4 from another of the numbers..., they all become equivalent (equal to 20).</span hide>