AMC 10A 2002 Problem 15 | Prime Number

Try this beautiful Problem based on Number theory from AMC 10A, 2002 Problem 15.

Prime Number | AMC 10A 2021, Problem 15

Using the digits $1,2,3,4,5,6,7$, and 9 , form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?

Key Concepts

Arithmetic

Divisibility

Prime Number

Suggested Book | Source | Answer

Elementary Number Theory by David M. Burton.

AMC 10A 2002 Problem 15

190

Try with Hints

First try to find the probable digits for the unit place of the prime number.

The two digit prime number should end with $1, 3, 7, 9$ since it is prime and should not divisible by $2$ or $5$.

So now try to find which two digit primes will work here.

So, the primes should be $23, 41, 59, 67$.

Now find the sum of them.

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Perfect square and Positive Integer | TOMATO B.Stat Objective 115

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Perfect square and Positive Integer.

Perfect Square & Positive Integers ( B.Stat Objective Question )

If n is a positive integer such that 8n+1 is a perfect square, then

  • n must be odd
  • 2n cannot be a perfect square
  • n must be a prime number
  • none of these

Key Concepts

Perfect square

Positive Integer

Primes

Check the Answer

Answer: 2n cannot be a perfect square.

B.Stat Objective Problem 115

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

Let \((8n+1)=k^{2}\) be a perfect square so k is found to be odd here as 8n+1 is odd.

\(\Rightarrow 8n=k^{2}-1\)

\(\Rightarrow 8n=(k-1)(k+1)\)

\(\Rightarrow 2n=\frac{(k-1)(k+1)}{4}\)

\(\Rightarrow (\frac{k-1}{2})(\frac{k+1}{2})\)

here (k-1) and (k+1) are consecutive even numbers then \((\frac{k-1}{2})(\frac{k+1}{2})\) are consecutive even numbers

\(2k \times (2k+2)= 2^2{k(k+1)}\) is not a perfect square as product of two consecutive numbers proved here as not a perfect square

So, 2n is not perfect square.

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GCD and Primes | PRMO 2017 | Question 29

Try this beautiful problem from the PRMO, 2017 based on GCD and Primes.

GCD and primes - PRMO 2017

For each positive integer n, consider the highest common factor \(h_n\) of the two numbers n!+1 and (n+1)! for n<100, find the largest value of \(h_n\).

  • is 107
  • is 97
  • is 840
  • cannot be determined from the given information

Key Concepts

GCD

Primes

Inequalities

Check the Answer

Answer: is 97.

PRMO, 2017, Question 29

Elementary Number Theory by David Burton

Try with Hints

n! +1 is not divisible by 1,2,.....,n (n+1)! divisible by 1,2,....,n then \(hcf \geq (n+1)\) and (n+1)! not divisible by n+2, n+3,...... then hcf= (n+1)

let n=99, 99! +1 and (100)! hcf=100 not possible as 100 |99! and 100 is non prime

let n=97 96! + 1 and 97! both divisible by 97 then hcf=97.

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