Difference between revisions of "1971 Canadian MO Problems/Problem 7"
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<cmath>1000a+100b+10d+e|9000a+900b+100c.</cmath> | <cmath>1000a+100b+10d+e|9000a+900b+100c.</cmath> | ||
− | Clearly we have that <math>8(1000a+100b+10d+e)<9000a+900b+100c<10(1000a+100b+10d+e)</math>, as <math>a</math> is positive. Therefore, this quotient must be equal to 9 (note that this does not mean n/m = 9), and | + | Clearly we have that <math>8(1000a+100b+10d+e)<9000a+900b+100c<10(1000a+100b+10d+e)</math>, as <math>a</math> is positive. Therefore, this quotient must be equal to 9 (note that this does not mean <math>n/m = 9</math>), and |
<cmath>9000a+900b+90d+9e=9000a+900b+100c.</cmath> | <cmath>9000a+900b+90d+9e=9000a+900b+100c.</cmath> |
Latest revision as of 15:47, 10 June 2020
Problem
Let be a five digit number (whose first digit is non-zero) and let be the four digit number formed from n by removing its middle digit. Determine all such that is an integer.
Solution
Let and , where , , , , and are base-10 digits and . If is an integer, then , or
This implies that
Clearly we have that , as is positive. Therefore, this quotient must be equal to 9 (note that this does not mean ), and
This simplifies to . The only way that this could happen is that . Then . Therefore the only values of such that is an integer are multiples of 1000. It is not hard to show that these are all acceptable values.
See Also
1971 Canadian MO (Problems) | ||
Preceded by Problem 6 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 8 |