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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AMC 10A 2020 Problem 6 | Divisibility Problem

Try this beautiful Problem based on Divisibility Problem from AMC 2020 Problem 6.

Divisibility Problem: AMC 10A 2020 Problem 6

How many 4-digit positive integers (that is, integers between 1000 and 9999 , inclusive) having only even digits are divisible by 5 ?

Key Concepts

Divisibility

Counting Principle

Suggested Book | Source | Answer

AMC 10A 2020 Problem 6

100

Try with Hints

What is the divisibility rule for a number divisible by 5?

Now apply this for unit, tens, hundred and thousand digits.

Here the unit digit must be 0. So I just have one choice for units place.

The middle two digits can be 0, 2, 4, 6, or 8.

But the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero.

Now try to count how many choices are there for each position.

Then there was 1 choice for unit digit.

5 choices for middle two digits.

4 choices for thousands digit.

Now calculate the total number of choices you can make.

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Math Kangaroo (Benjamin) 2020 Problem 26 | Divisibility Rule

Try this beautiful Problem based on Divisibility Rule from Math Kangaroo (Benjamin) 2020.

Math Kangaroo (Benjamin), 2020 | Problem No 26

We say that a three-digit number is balanced if the middle digit is the arithmetic mean of the other two digits. How many balanced numbers are divisible by 18 ?

Key Concepts

Arithmetic

Divisibility Rule

Arithmetic Mean

Suggested Book | Source | Answer

Algebra by Gelfand

Math Kangaroo (Benjamin), 2020

6

Try with Hints

Let the number be $abc.$

By the given condition of arithmetic mean we get,

$$\frac{a+c}{2}=b$$.

Now from here can I conclude $a, c$ are of the same parity?

Given that the number is divisible by $18$

$\Rightarrow$ Given number must be divisible by both $2$ and $9$.

Apply divisibility rule of $9$ to guess the values of $a, b, c$.

Here the values of $3b$ are $9, 18, 27$.

$\Rightarrow$ the values of $2b$ are $6, 12, 18$.

Now guess the values of $a+c$?

And then find the possible pair of $(a, c)$?

The possible values of $(a, c)$ are $(2,4)$, $(4,2)$, $(6,0)$, $(4,8)$, $(6,6)$, $(8,4)$.

So, how many possible numbers are there?

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