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Positive Integer | PRMO-2017 | Question 1

Try this beautiful Positive Integer Problem from Algebra from PRMO 2017, Question 1.

Positive Integer - PRMO 2017, Question 1


How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3 ?

  • 9
  • 7
  • 28

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:28

PRMO-2017, Problem 1

Pre College Mathematics

Try with Hints


Let n be the positive integer less than 1000 and s be the sum of its digits, then 3 \mid n and 7 \mid s
3|n \Rightarrow 3| s
therefore21| s

Can you now finish the problem ..........

Also n<1000 \Rightarrow s \leq 27
therefore \mathrm{s}=21
Clearly, n must be a 3 digit number Let x_{1}, x_{2}, x_{3} be the digits, then x_{1}+x_{2}+x_{3}=21
where 1 \leq x_{1} \leq 9,0 \leq x_{2}, x_{3} \leq 9
\Rightarrow x_{2}+x_{3}=21-x_{1} \leq 18
\Rightarrow x_{1} \geq 3

Can you finish the problem........

For x_{1}=3,4, \ldots ., 9, the equation (1) has 1,2,3, \ldots ., 7 solutions
therefore total possible solution of equation (1)

=1+2+\ldots+7=\frac{7 \times 8}{2}=28

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Try this beautiful Positive Integer Problem from Algebra from PRMO 2017, Question 1.

Positive Integer - PRMO 2017, Question 1


How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3 ?

  • 9
  • 7
  • 28

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:28

PRMO-2017, Problem 1

Pre College Mathematics

Try with Hints


Let n be the positive integer less than 1000 and s be the sum of its digits, then 3 \mid n and 7 \mid s
3|n \Rightarrow 3| s
therefore21| s

Can you now finish the problem ..........

Also n<1000 \Rightarrow s \leq 27
therefore \mathrm{s}=21
Clearly, n must be a 3 digit number Let x_{1}, x_{2}, x_{3} be the digits, then x_{1}+x_{2}+x_{3}=21
where 1 \leq x_{1} \leq 9,0 \leq x_{2}, x_{3} \leq 9
\Rightarrow x_{2}+x_{3}=21-x_{1} \leq 18
\Rightarrow x_{1} \geq 3

Can you finish the problem........

For x_{1}=3,4, \ldots ., 9, the equation (1) has 1,2,3, \ldots ., 7 solutions
therefore total possible solution of equation (1)

=1+2+\ldots+7=\frac{7 \times 8}{2}=28

Subscribe to Cheenta at Youtube


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