Answer
424.5k+ views
Hint:One approach to the question is for you to make up some simple number sentences in the format “number +/–/x/÷ number = number” and then to look for a repeated digit that they could remove and replace with a alphabet.
You need to think about patterns (or strategies) that you may use, such as:
• odd + odd = even (for example: 9 + 5 = 14)
• odd – odd = even (for example: 9 – 5 = 4)
• even + even = even
• even + odd = odd
• even – odd = odd
• odd – even = odd
• odd x odd = odd
• even x even = even
• odd x even = even.
Complete step-by-step answer:
If ′A′ stands for 2 and ′C′ stands for 6
When ‘A’, ‘B’ and ‘C’ are multiplied by ‘E’ then the product doesn’t change. So, the value of ‘E’ = 1
So, now by replacing the value of ‘A’, ‘C’ and ‘E’
When ‘2B6’ is multiplied with ‘D’ then 2 comes in unit place.
Now, 2 appears as a unit digit when 6 is multiplied by either 2 or 7.
Value of ‘D’ cannot be 2 as 2 is already a value of ‘A’.
So value of ‘D’ =7
As we can see out of 10 integers we have already used 1, 2, 6, 7
So, now we are left with 0, 3, 4, 5, 8, 9
So, lets try to put the value of ‘B’ as 0
The calculation is not coming correct.
So, lets try to put the value of ‘B’ as 3
The calculation is coming correct.
So, the value of ‘M’ is 5
So, the correct answer is “Option C”.
Note:Most of these types of questions could be done using trial and improvement, but you will like to find it more efficient (and more satisfying) to first narrow the range of options as much as possible by studying the problem and seeing what you can deduce.We should be careful while choosing numbers as in question said that all letters represents different integers so none of them should not be repeated.
You need to think about patterns (or strategies) that you may use, such as:
• odd + odd = even (for example: 9 + 5 = 14)
• odd – odd = even (for example: 9 – 5 = 4)
• even + even = even
• even + odd = odd
• even – odd = odd
• odd – even = odd
• odd x odd = odd
• even x even = even
• odd x even = even.
Complete step-by-step answer:
If ′A′ stands for 2 and ′C′ stands for 6
When ‘A’, ‘B’ and ‘C’ are multiplied by ‘E’ then the product doesn’t change. So, the value of ‘E’ = 1
So, now by replacing the value of ‘A’, ‘C’ and ‘E’
![seo images](https://www.vedantu.com/question-sets/afda6246-3709-4b21-b573-54346b9833d9311760985159537983.png)
When ‘2B6’ is multiplied with ‘D’ then 2 comes in unit place.
Now, 2 appears as a unit digit when 6 is multiplied by either 2 or 7.
Value of ‘D’ cannot be 2 as 2 is already a value of ‘A’.
So value of ‘D’ =7
As we can see out of 10 integers we have already used 1, 2, 6, 7
So, now we are left with 0, 3, 4, 5, 8, 9
So, lets try to put the value of ‘B’ as 0
![seo images](https://www.vedantu.com/question-sets/f803a68a-1d32-4cc6-b397-061f56b4f9a75987420985348494967.png)
The calculation is not coming correct.
So, lets try to put the value of ‘B’ as 3
![seo images](https://www.vedantu.com/question-sets/ad9cfc3c-c486-4b1c-92f5-acd6251a84dd3926803011772145211.png)
The calculation is coming correct.
So, the value of ‘M’ is 5
So, the correct answer is “Option C”.
Note:Most of these types of questions could be done using trial and improvement, but you will like to find it more efficient (and more satisfying) to first narrow the range of options as much as possible by studying the problem and seeing what you can deduce.We should be careful while choosing numbers as in question said that all letters represents different integers so none of them should not be repeated.
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