Difference between revisions of "1997 AJHSME Problems/Problem 23"
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There are positive integers that have these properties: | There are positive integers that have these properties: | ||
− | * | + | *The sum of the squares of their digits is equal to 50 |
− | * | + | *Each digit is larger than the one on it's left |
The product of the digits of the largest integer with both properties is | The product of the digits of the largest integer with both properties is | ||
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==Solution== | ==Solution== | ||
− | Five digit numbers will have a minimum of <math>1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55</math> as the sum of their squares if the five digits are distinct and non-zero. If there is a zero, it will be forced to the left by rule #2. | + | Five-digit numbers will have a minimum of <math>1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55</math> as the sum of their squares if the five digits are distinct and non-zero. If there is a zero, it will be forced to the left by rule #2. |
No digit will be greater than <math>7</math>, as <math>8^2 = 64</math>. | No digit will be greater than <math>7</math>, as <math>8^2 = 64</math>. | ||
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<math>z=7</math> will not work, since the other digits must be at least <math>1^2 + 2^2 + 3^2 = 14</math>, and the sum of the squares would be over <math>50</math>. | <math>z=7</math> will not work, since the other digits must be at least <math>1^2 + 2^2 + 3^2 = 14</math>, and the sum of the squares would be over <math>50</math>. | ||
− | <math>z=6</math> will give <math>w^2 + x^2 + y^2 = 14</math>. <math>(w,x,y) = (1,2,3)</math> will work, giving the number <math>1236</math>. No other number with <math>z=6</math> will work, as <math>w, x, </math> and <math>y</math> would | + | <math>z=6</math> will give <math>w^2 + x^2 + y^2 = 14</math>. <math>(w,x,y) = (1,2,3)</math> will work, giving the number <math>1236</math>. No other number with <math>z=6</math> will work, as <math>w, x, </math> and <math>y</math> would have to be greater. |
− | <math>z=5</math> will give <math>w^2 + x^2 + y^2 = 25</math>. <math>y=4</math> forces <math>x=3</math> and <math>w=0</math>, which has a leading zero, and then we have 345 which is 3-digit number. | + | <math>z=5</math> will give <math>w^2 + x^2 + y^2 = 25</math>. <math>y=4</math> forces <math>x=3</math> and <math>w=0</math>, which has a leading zero, and then we have 345 which is a 3-digit number. |
<math>z=4</math> can only give the number <math>1234</math>, which does not satisfy the condition of the problem. | <math>z=4</math> can only give the number <math>1234</math>, which does not satisfy the condition of the problem. |
Latest revision as of 11:07, 24 December 2024
Problem
There are positive integers that have these properties:
- The sum of the squares of their digits is equal to 50
- Each digit is larger than the one on it's left
The product of the digits of the largest integer with both properties is
Solution
Five-digit numbers will have a minimum of as the sum of their squares if the five digits are distinct and non-zero. If there is a zero, it will be forced to the left by rule #2.
No digit will be greater than , as .
Trying four digit numbers , we have with
will not work, since the other digits must be at least , and the sum of the squares would be over .
will give . will work, giving the number . No other number with will work, as and would have to be greater.
will give . forces and , which has a leading zero, and then we have 345 which is a 3-digit number.
can only give the number , which does not satisfy the condition of the problem.
Thus, the number in question is , and the product of the digits is , giving as the answer.
See also
1997 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.