Question
Find the digit A
B A×B 3 57A
B A×B 3 57A
Answer: Option C
:
C
Ones digit of 3 × A is A. So, it must be either A = 0 or A = 5.
Now, if B = 1, then BA ×B3 would at most be equal to 15× 13; (Taking A= 5 to find the maximum value of the product)that is,it would at most be equal to 195. But the product here is 57A, which is more than 500. So we cannot have B =1.
If B = 3, then BA × B3 would be at least30× 33; (Taking A = 0 to find the minimum value of the product) that is990. But 57A is less than 990. So, B cannot be equal to 3.
Putting these two facts together, we get B = 2.
So, the multiplication is either 20 × 23, or
25 × 23.
The first possibility fails, since
20 × 23 = 460. But, the second one works out correctly, since
25 × 23 = 575. So the answer is A = 5 andB = 2.
25×23575
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:
C
Ones digit of 3 × A is A. So, it must be either A = 0 or A = 5.
Now, if B = 1, then BA ×B3 would at most be equal to 15× 13; (Taking A= 5 to find the maximum value of the product)that is,it would at most be equal to 195. But the product here is 57A, which is more than 500. So we cannot have B =1.
If B = 3, then BA × B3 would be at least30× 33; (Taking A = 0 to find the minimum value of the product) that is990. But 57A is less than 990. So, B cannot be equal to 3.
Putting these two facts together, we get B = 2.
So, the multiplication is either 20 × 23, or
25 × 23.
The first possibility fails, since
20 × 23 = 460. But, the second one works out correctly, since
25 × 23 = 575. So the answer is A = 5 andB = 2.
25×23575
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