Difference between revisions of "1980 AHSME Problems/Problem 30"
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== Solution == | == Solution == | ||
| − | < | + | N = a^2*10000 + b^2*100 + c^2*1 |
| + | = (a*100 + c)^2 | ||
| + | |||
| + | we get | ||
| + | b^2 = 2*a*c | ||
| + | |||
| + | where | ||
| + | 4<=a,b,c<=9 | ||
| + | |||
| + | |||
| + | b^2=2*a*c, so | ||
| + | |||
| + | a=2*2, c=2*2*2, b=8 | ||
| + | a=2*3, c=3 (NO) | ||
| + | a=2*2*2, c=3*3, b=3*4=12 (NO) | ||
| + | |||
| + | Therefore, there are two solutions (4,8,8) or (8,8,4) | ||
| + | |||
| + | [[User:Wwei.yu|Wwei.yu]] ([[User talk:Wwei.yu|talk]])Wei | ||
== See also == | == See also == | ||
Latest revision as of 21:49, 28 March 2020
Problem
A six digit number (base 10) is squarish if it satisfies the following conditions:
(i) none of its digits are zero;
(ii) it is a perfect square; and
(iii) the first of two digits, the middle two digits and the last two digits of the number are all perfect squares when considered as two digit numbers.
How many squarish numbers are there?
Solution
N = a^2*10000 + b^2*100 + c^2*1
= (a*100 + c)^2
we get
b^2 = 2*a*c
where 4<=a,b,c<=9
b^2=2*a*c, so
a=2*2, c=2*2*2, b=8 a=2*3, c=3 (NO) a=2*2*2, c=3*3, b=3*4=12 (NO)
Therefore, there are two solutions (4,8,8) or (8,8,4)
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Problem 30 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
