Observe the pattern : 1 × 1 = 1 11 × 11 = 121 111 × 111 = 12321 1111 × 1111 = ________ 11111 × 11111 = ________

Given, 1 × 1 = 1 11 × 11 = 121 111 × 111 = 12321 1111 × 1111 = ________ 11111 × 11111 = ________ First of all, number of digits of the multiplied numbers are all odd. Since digit is odd. if we pick the middle number, each side will contain even number of digits: and those numbers are in reversed symmetrical order. 1 × 1 = 1 11 × 11 = 121 111 × 111 = 12321 1111 × 1111 = 1234321 11111 × 11111 = 123454321

Observe the pattern and extend it to four more steps: 7 × 2 - 3 = 11 7 × 3 - 6 = 15 7 × 4 - 9 = 19 7 × 5 - 12 = 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Given pattern: 7 × 2 - 3 = 11 7 × 3 - 6 = 15 7 × 4 - 9 = 19 7 × 5 - 12 = 23 By observing the above pattern, we find that in each step, the multiplicand 7 is multiplied by the numbers 2 , 3 , 4 , 5 , . . . . . and are subtracted by 3 , 6 , 9 , 12 ∵ 0 + 3 , 6 ∵ 3 + 3 , 9 ∵ 6 + 3 , . . . . . . respectively. By following the above pattern, the next steps are as follows: ⇒ 7 × 6 - 15 = 27 ⇒ 7 × 7 - 18 = 31 ⇒ 7 × 8 - 21 = 35 ⇒ 7 × 9 - 24 = 39 Hence, the required solution is obtained.

Observe the following pattern: 8 × 1 + 1 = 9 8 × 3 + 1 = 25 8 × 6 + 1 = 49 8 × 10 + 1 = 81 write three more steps.

Given pattern: 8 × 1 + 1 = 9 8 × 3 + 1 = 25 8 × 6 + 1 = 49 8 × 10 + 1 = 81 By observing the above pattern, we find that in each step, the multiplicand 8 is multiplied by the numbers 1 , 3 , 6 , 10 , . . . . . and are added by 1 , 9 ∵ 8 + 1 , 25 ∵ 24 + 1 , 49 ∵ 48 + 1 , . . . . . . respectively. By following the above pattern, the next steps are as follows: ⇒ 8 × 15 + 1 = 121 ⇒ 8 × 21 + 1 = 169 ⇒ 8 × 28 + 1 = 225 Hence, the required solution is obtained.

Observe the following pattern: 1 × 8 + 1 = 9 12 × 8 + 2 = 98 123 × 8 + 3 = 987 write two more steps.

Given that, 1 × 8 + 1 = 9 12 × 8 + 2 = 98 123 × 8 + 3 = 987 If we observe the above pattern clearly, in R.H.S the units place is one number less than the number in the units place of its preceding number and the digits left to units place is same number as preceding digit. So, the next step is as follows 1234 × 8 + 4 = 9876 12345 × 8 + 5 = 98765 Hence, the required solution is obtained.

E M B I B E

Explore Mathematics 6>Chapter 2 - Whole Numbers>EXERCISE 2.6

Jagat Jitendra Matto Solutions for Chapter: Whole Numbers, Exercise 7: EXERCISE 2.6

Author:Jagat Jitendra Matto

Jagat Jitendra Matto Mathematics Solutions for Exercise - Jagat Jitendra Matto Solutions for Chapter: Whole Numbers, Exercise 7: EXERCISE 2.6

Attempt the free practice questions on Chapter 2: Whole Numbers, Exercise 7: EXERCISE 2.6 with hints and solutions to strengthen your understanding. Explore Mathematics 6 solutions are prepared by Experienced Embibe Experts.

Questions from Jagat Jitendra Matto Solutions for Chapter: Whole Numbers, Exercise 7: EXERCISE 2.6 with Hints & Solutions

HARD

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 2 - Whole Numbers>EXERCISE 2.6>Q 1

Observe the pattern :

$1 \times 1 = 1 11 \times 11 = 121 111 \times 111 = 12321 1111 \times 1111 =________11111 \times 11111 =________$

HARD

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 2 - Whole Numbers>EXERCISE 2.6>Q 2

Observe the pattern and extend it to four more steps:

$(7 \times 2) - 3 = 11 (7 \times 3) - 6 = 15 (7 \times 4) - 9 = 19 (7 \times 5) - 12 = 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$