99*54 there is something tricky
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99×54 is 5346 I don't think that there's something tricky
Priyanshupandey11:
you are right that was just to confuse you
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the answer is 5346
Multiplication by 9, 99, 999, etc.
One way to multiply a number by 9 is to multiply by 10 and then subtract the number from the product. There is another way to multiply fast by 9 and as the first one it has an analogue for multiplication by 99, 999 and all such numbers. Let's start with the multiplication by 9.
To multiply a one digit number a by 9, first subtract 1 and form b = a - 1. Next, subtract b from 9: c = 9 - b. Then just write b and c next to each other:
9a = bc.
For example, find 6×9 (so that a = 6.) First subtract: 5 = 6 - 1. Subract the second time: 4 = 9 - 5. Lastly, form the product 6×9 = 54.
Next, find 37×99. First, subtract 1: 36 = 37 - 1. Then subtract 63 = 99 - 36. Lastly, form the product: 37×99 = 3663.
Why does this work? For the multiplication by 9, bc= 10b + c:
bc= 10b + c = 10(a - 1) + (9 - (a - 1)) = 10a - 10 + 10 - a = 9a,
as required. Similarly, for a 2-digit a:
bc= 100b + c = 100(a - 1) + (99 - (a - 1)) = 100a - 100 + 100 - a = 99a.
Do try the same derivation for a three digit number. As an example,
543×999= 1000×542 + (999 - 542) = 999×542 + 999 = 999×543
just by using the distributive law twice
Multiplication by 9, 99, 999, etc.
One way to multiply a number by 9 is to multiply by 10 and then subtract the number from the product. There is another way to multiply fast by 9 and as the first one it has an analogue for multiplication by 99, 999 and all such numbers. Let's start with the multiplication by 9.
To multiply a one digit number a by 9, first subtract 1 and form b = a - 1. Next, subtract b from 9: c = 9 - b. Then just write b and c next to each other:
9a = bc.
For example, find 6×9 (so that a = 6.) First subtract: 5 = 6 - 1. Subract the second time: 4 = 9 - 5. Lastly, form the product 6×9 = 54.
Next, find 37×99. First, subtract 1: 36 = 37 - 1. Then subtract 63 = 99 - 36. Lastly, form the product: 37×99 = 3663.
Why does this work? For the multiplication by 9, bc= 10b + c:
bc= 10b + c = 10(a - 1) + (9 - (a - 1)) = 10a - 10 + 10 - a = 9a,
as required. Similarly, for a 2-digit a:
bc= 100b + c = 100(a - 1) + (99 - (a - 1)) = 100a - 100 + 100 - a = 99a.
Do try the same derivation for a three digit number. As an example,
543×999= 1000×542 + (999 - 542) = 999×542 + 999 = 999×543
just by using the distributive law twice
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