Squaring numbers near 50 and 100 and the ones ending with 5 TRICKS

Question

Squaring numbers ending with 5

Suppose you want to find the square of 85

write 5^2 = _ _25
and then 8*9 = 72 i.e the digit before 5 multiplied by next consecutive integer
so, 85^2 = 7225

similarly,

Suppose you want to find the square of 115
write 5^2 = _ _25
and then 11*12 = 132 i.e the no. before 5 multiplied by next consecutive integer
so, 115^2 = 13225

Squaring numbers near 100

108^2 = ??

take d = (number-100)^2 and this will be the last two digits of the answer
i.e. (108-100)^2 = 8^2 = 64 Now, the number is your base.
number + |d| , this will be your first three digits
108 + 8 = 116

108^2 = 11664

94^2 = ??

take d = (number-100)^2 and these will be the last two digits of the answer
i.e. (94-100)^2 = 6^2 = 36 Now, the number is your base.
number - |d| , this will be your first two digits
94 - 6 = 88

94^2 = 8836

These will work for no's near 50 as well but with a different base in the second step. Here your base will be 25.

47^2 = ??

Take d = (number-50)^2 and this will be the last two digits of the answer
i.e. (47-50)^2 = 3^2 = 09
Now instead of the no. you have to take 25 as base.
25 - |d| , this will be your first two digits
25 - 3 = 22

47^2 = 2209

58^2 = ??

Take d = (number-50)^2 and this will be the last two digits of the answer
i.e. (58-50)^2 = 8^2 = 64
Now instead of the no. you have to take 25 as base.
25 + |d| , this will be your first two digits
25 + 8 = 33

58^2 = 3364

You will be able to square any no. in the range in less than 10 seconds.
Best of luck. Kudos if helpful.

IanStewart · Answer

You can often use the difference of squares pattern, a^2 - b^2 = (a+b)(a-b), to calculate squares quickly. We just need to subtract a square so that at least one of the factors (a+b) or (a-b) will be easy to work with. So if we wanted to compute the value of 85^2, we could set up an equation where we subtract 5^2 on the left side, use the difference of squares factorization on the right side, then solve for 85^2:

\begin{align}
85^2 - 5^2 &= (85 + 5)(85-5) \
85^2 - 25 &= (90)(80) \
85^2 &= 7200 + 25 = 7225
\end{align}

We choose to subtract 5^2 above because then the difference of squares factorization will give us two simple numbers to multiply. Subtracting 15^2 would also work out cleanly. Similarly, if we want to calculate the value of 94^2, we can subtract 6^2. Then the (a+b) factor ends up being 100, which is easy to multiply by:

\begin{align}
94^2 - 6^2 &= (94+6)(94-6) \
94^2 - 36 &= (100(88) \
94^2 &= 8800 + 36 = 8836
\end{align}

One advantage to using the difference of squares here is that you don't need to learn different techniques for different types of numbers. We can use it for some three-digit numbers as well, so if we want to compute 997^2, we can subtract 3^2 first:

\begin{align}
997^2 - 3^2 &= (997 + 3)(997 - 3) \
997^2 - 9 &= (1000)(994) \
997^2 &= 994000 + 9 = 994009 
\end{align}

But there's a more important advantage for GMAT test takers. While you almost never need to calculate awkward squares on the GMAT, you almost always need to use the difference of squares somewhere. And that's the main reason I like the approach above - even if a test taker never needs to use it to actually compute a square, practicing it may help him or her to recognize opportunities to use the difference of squares pattern in the kinds of questions that do appear on the GMAT.

bumpbot · Answer

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Squaring numbers near 50 and 100 and the ones ending with 5 TRICKS

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