Divisibility Problems

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Calendar Problems Problems on LCM

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Problems on divisibility rules will help us to learn how to use the rules to test of divisibility by 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11.

divisibility problems shortcuts

Divisibility by 2

Rule: The last digit which is either even or 0, is divisible by 2. Example 34, 2532, 1290

Divisibility by 3

Rule: If the sum of the digits of a number is divisible by 3, then the number is also divisible by 3 Example 2451, so required sum 2+4+5+1=12 divisible by 3 so the number 2451 is also divisible by 3

Divisibility by 4

Rule 1: If the last two digits of a number is divisible by 4, the number is also divisible by 4

Rule 2: The number having two or more zeros at the end is also divisible by 4

Example 728524 since the last two digits are divisible by 4 so the number is also divisible by 4. Example 15600 last two digits are 00 so the number is also divisible by 4.

Divisibility by 5

Rule: If a number ends with 5, 0 then the number will be divisible by 5 Example 1765, 12330

Divisibility by 6

For this both two rule should be full filled

Rule 1: The number should end with an even digit or 0.

Rule 2: The sum of the digits should divisible by 3.

Example 174 is divisible by 6 as the number ends with even digit 4 and sum of the digits 12 is divisible by 3

Example 8520 is divisible by 6, as it ends with 0 and sum of the number 15 is divisible by 3

Divisibility by 7

There is no any stritct rule for it

Example 896 Solution: 896 89-6*2=77 as 77 is divisible to 7 so this number 896 is also divisible to 7

Example 4753 Solution: 475-3*2= 469 46-9*2=28 as 28 is divisible by 7 so the number 4753 is divisible to 7

Divisibility by 8

Rule 1: If the last three digits of the number is divisible by 8 , the number is also divisible by 8

Rule 2: If the last three digits of the number is three zeros 000 then the number is also divisible to 8

Example 14873258376256 find out divisibility by 8. Solution: 14873258376256 is divisible to 8 , so the whole number is divisible to 8

Divisibility by 9

Rule: If the sum of all the digits of a number is divisible by 9, hence the number is also divisible by 9. Example 8758323 Solution: Sum of the number is 8+7+5+8+3+2+3=36 which is divisible by 9, so the number is divisible by 9

Divisibility by 11

Rule: If the sum of the digits at odd places and even places are equal or differ by an amount of 11, then the number will be divisible by 11

Example 589743671 Check the divisibility by 11 Solution: Sum of the digits at odd places 5+9+4+6+1=25 Sum of the digits at even places 8+7+3+7=25 So the number is divisible by 11

Example 9754239 check the divisibility by 11 Solution: Sum of the numbers at odd places 9+5+2+9=25 Sum of the numbers at even places 7+4+3=14, the difference 25-14=11, so the number will be divisible by 11

Divisibility by 12

Rule: Any number which is divisible by both 4 and 3, is also divisible by 12

Divisibility by 14

Rule: for divisibility with 14, the number should be even and should be divisible by 7

Divisibility by 15

Rule: Any number which is divisible by both 3 and 5 is also divisible by 15

Divisibility by 16

Rule: any number whose last four digits number is divisible by 16 is also divisible by 16

Question 1: Which of the following numbers is divisible by 3?

  • (A) 51412
  • (B) 86221
  • (C) 76322
  • (D) 63693

Answer: (D) 63693

Solution

For, 51412 (5+1+4+1+2) = 13, which is NOT divisible by 3. Hence 51412 is not divisible by 3.
For, 86221 (8+6+2+2+1) = 19, which is NOT divisible by 3. Hence 86221 is not divisible by 3.
For, 76322 (7+6+3+2+2) = 20, which is NOT divisible by 3. Hence 76322 is not divisible by 3.
For, 63693 (6+3+6+9+3) = 27, which IS divisible by 3. Hence 63693 is divisible by 3.

Question 2: When the integer n is divided by 8, the remainder is 3. What is the remainder if 6n is divided by 8?

  • (A) 0
  • (B) 1
  • (C) 2
  • (D) 3
  • (E) 4

Answer: (C) 2

Solution

When n is divided by 8, the remainder is 3 may be written as

n = 8 k + 3

multiply all terms by 6

6 n = 6(8 k + 3) = 8(6k) + 18

Write 18 as 16 + 2 since 16 = 8 * 2.

= 8(6k) + 16 + 2

Factor 8 out.

= 8(6k + 2) + 2

The above indicates that if 6n is divided by 8, the remainder is 2. The answer is C.

Question 3: Which one of the following is not a prime number?

  • (A) 31
  • (B) 61
  • (C) 71
  • (D) 91

Answer: (D) 91

Explanation: 91 is divisible by 7. So, it is not a prime number.

Question 4: 1397 x 1397 = ?

  • (A) 1951609
  • (B) 1981709
  • (C) 18362619
  • (D) 2031719
  • (E) None of these

Answer: (A) 1951609

Solution

1397 x 1397	= (1397)2
= (1400 - 3)2
= (1400)2 + (3)2 - (2 x 1400 x 3)
= 1960000 + 9 - 8400
= 1960009 - 8400
= 1951609.

Question 5: 5358 x 51 = ?

  • (A) 273258
  • (B) 273268
  • (C) 273348
  • (D) 273358

Answer: (A) 273258

Solution

5358 x 51	= 5358 x (50 + 1)
= 5358 x 50 + 5358 x 1
= 267900 + 5358
= 273258.

Question 6: The sum of first five prime numbers is:

  • (A) 11
  • (B) 18
  • (C) 26
  • (D) 28

Answer: (D) 28

Solution

Required sum = (2 + 3 + 5 + 7 + 11) = 28.

Note: 1 is not a prime number.

Definition:

A prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 1 and itself.

Question 7: If the number 517*324 is completely divisible by 3, then the smallest whole number in the place of * will be:

  • (A) 0
  • (B) 1
  • (C) 2
  • (D) None of these

Answer: (C) 2

Solution

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = (22 + x), which must be divisible by 3.

x = 2.

Calendar Problems Problems on LCM

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