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Difference between revisions of "1998 CEMC Gauss (Grade 7) Problems/Problem 25"

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If the product <math>pq</math> is a power of <math>10,</math> and both <math>p</math> and <math>q</math> do not end in 0, then <math>p</math> must be in the form <math>5^n</math> and <math>q</math> must be in the form <math>2^n.</math>
 
If the product <math>pq</math> is a power of <math>10,</math> and both <math>p</math> and <math>q</math> do not end in 0, then <math>p</math> must be in the form <math>5^n</math> and <math>q</math> must be in the form <math>2^n.</math>
  
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Start looking at small values of <math>n</math> and subtract:
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We know that <math>5^n \equiv 5 \pmod {10}</math> for all positive integers <math>n</math> and <math>2^n \not\equiv 0 \pmod {10}</math> for all integers <math>n</math>.
  
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<cmath>p=5, \quad q=2, \quad \rightarrow \quad p-q \equiv 3</cmath>
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Therefore, we know that <math>p - q \not\equiv 5 - 0 = \boxed{\textbf{(C)}\ 5}.</math>
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<cmath>p=25, \quad q=4, \quad \rightarrow \quad p-q \equiv 1</cmath>
 
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<cmath>p=125, \quad q=8, \quad \rightarrow \quad p-q \equiv 7</cmath>
 
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<cmath>p=625, \quad q=16, \quad \rightarrow \quad p-q \equiv 9</cmath>
 
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This pattern continues in groups of <math>4</math>, and the only number not included is <math>\boxed{\textbf{(C)}\ 5}.</math>
 
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-edited by coolmath34
 

Latest revision as of 17:59, 7 April 2024

Problem

Two natural numbers, $p$ and $q,$ do not end in zero. The product of any pair, $p$ and $q,$ is a power of 10 (that is, 10, 100, 1000, 10 000 , ...). If $p > q$ , the last digit of $p-q$ cannot be

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9$

Solution

If the product $pq$ is a power of $10,$ and both $p$ and $q$ do not end in 0, then $p$ must be in the form $5^n$ and $q$ must be in the form $2^n.$

We know that $5^n \equiv 5 \pmod {10}$ for all positive integers $n$ and $2^n \not\equiv 0 \pmod {10}$ for all integers $n$.

Therefore, we know that $p - q \not\equiv 5 - 0 = \boxed{\textbf{(C)}\ 5}.$

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