Difference between revisions of "1997 AJHSME Problems/Problem 23"
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− | 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 | + | <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 15:15, 6 May 2023
Problem
There are positive integers that have these properties:
- the sum of the squares of their digits is 50, and
- each digit is larger than the one to its 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.