Math, asked by nikshay2, 1 year ago

4. Find out (A+B+C+D) such that AB ×CB =DDD.where
AB and CB are two-digit numbers and DDD is a three-
digit number.
(A) 21
(B) 19
(C) 17
(D) 18

Answers

Answered by haifamaryamaqeel
1

Answer:

First concentrate on units. B(squared) should end with D. D≠B

Step-by-step explanation:

Why am I taking B as B squared? Because we are multiplying B by itself in AB X CB = DDD.

Possible values of D is 1,4,6 or 9

111=3∗37

444=2∗2∗3∗37  

666=2∗3∗3∗37

999=3∗3∗3∗37=27∗37

Only one possibility is seen = 27∗37=999

In this answer, AB and CB are two digit numbers, and DDD is a three digit number.

Hope this helps :)

Answered by gangadharpatila123
0

Answer:

Step-by-step explanation:

Primary SchoolMath 5+3 pts

4. Find out (A+B+C+D) such that AB ×CB =DDD.where

AB and CB are two-digit numbers and DDD is a three-

digit number.

(A) 21

(B) 19

(C) 17

(D) 18

Ask for details FollowReport by Nikshay2 5 hours ago

Answers

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Gangadharpatila123 · Ambitious

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haifamaryamaqeel

Haifamaryamaqeel Ambitious

Answer:

First concentrate on units. B(squared) should end with D. D≠B

Step-by-step explanation:

Why am I taking B as B squared? Because we are multiplying B by itself in AB X CB = DDD.

Possible values of D is 1,4,6 or 9

111=3∗37

444=2∗2∗3∗37

666=2∗3∗3∗37

999=3∗3∗3∗37=27∗37

Only one possibility is seen = 27∗37=999

In this answer, AB and CB are two digit numbers, and DDD is a three digit number.

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