Nice! thanks a lot! Kudos

+1 kudo!
Great post dude!
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I started using the DS strategy after attended a test session from Manhattan GMAT. It GREATLY improved the accuracy of my DS answers. Great summary!

Ok guys! I couldn't find how to take square of any number especially 2 digit numbers and I think instead of cramming squares of numbers up to 25 why not find a fast and easy way to get the square (power 2) of any number. By the way I'll update this in GMAT Quant Notes file posted in first post. And yea I'll prove this method as well
Let's start from easy one, (P.S. The beauty of math is that you can solve it in several ways)
The formula is as follows
(a-d)^2= a^2 + d^2 -2ad
which can also be written as
(a-d)^2=a^2 - 2ad + d^2
Now by taking "a" as common in underlined portion of the equation, we get
=>a*(a-2d) + d^2
Now remember this one and let's take Square of 7 (start from easy one). To take square of & we'll use above formula and the rule is always to get to nearest number multiple of 10 meaning number with "0" as unit digit or number with the base 10 . Here in case it is "10" so to get to 10 we have to add "3" to "7", which means
7= 10-3 --> same as (a-d), so "10" (the base number) is as "a" and "3" is as "d", the difference between "7" and nearest number that is multiple of 10.
Now let's solve it
7^2= (10-3)^2= [10*{10-2*(3)}] + (3)^2={10*(10-6)} + 9 = (10*4)+ 9=40+9=49 proved

Whenever the square of any number comes in your mind like in this case "7", quickly add some number(or integer) to round it to a nearest number that is multiple of 10(the one with base 10) like in this it would be "10" so the difference is "3" now similarly subtract the difference from "7" which would be 7-3=4, Now multiple 10 (base 10 number) and 4 (after subtracting the difference from actual number) and then add the square of difference which would yield as 3^2=9. now let's look at this

To take the square of 7 let's first get the difference between base 10 number and 7, which in this case is 10-7=3=d (difference), Suppose a=7 (actual number whose square is to be taken) and d=3 (difference between actual number and base 10 number), Now let's get the formula and solve it
a^2= (a-d)*(a+d) + d^2=
by solving underlined portion to (a^2 - d^2) and the put it in above equation we get
=> (a^2 - d^2) + d^2= a^2 - d^2 + d^2
now the terms in underlined portion will cancel each other and we'll get only
=> a^2 Hence proved (see above the bold term a^2 which is equal to this term)
Now solving for 7
7^2=(7+3)*(7-3)+ 3^2=10*4 + 9=40+9=49
We can also use (a+d)^2, this is used when the number to be squared is just above Base number (multiple of 10 or 100 and so on)
And (a-d)^2 is used for the number just below the base number (multiple of 10 or 100 and so on)
Now for (a+d)^= a^2 + d^2 + 2ad= a^2 + 2ad + d^2= a*(a+2d) + d^2
Let's take 11
11^2= (10+1)^2 = 10 (10 + 2*1) + 1^2= 10 (10+2) +1= 10*(12) + 1= 120 +1=121 proved

Another way to solve it is (2nd method mentioned above already)
11^2= (11-1)*(11+1) + 1^2= (10*12) + 1= 120 +1=121
it's fast and easy once you get a grasp over it, multiplying any number with a number with base 10 makes it easy as you have to just put "0"s on right side of the resultant number and then add the square of difference gives the square of the number.
Now let's practice with different numbers,
68^2= 70*(70- 2*2) + 2^2= 70*(70-4) + 4 = 70*66 + 4= 4620 + 4= 4624
by the way multiplying underlined portion is easy as (by breaking down the numbers)
70*66=70*(60+6)=(70*60) + (70*6)=4200 + 420 = 4200+400+20=4600+20=4620 and then add 4 gives 4624=68^2
similarly bold portion of above equation can also be solved as
70*(70-4)= (70*70) - (70*4)=4900 - 280=4900 -200-80=4700-80=4620 and then adding 4 gives 4624=68^2
Now with other way
68^2= (68+2)*(68-2) + 2^2=70*66 + 2^2=4620+4=4624
Now let's take 46
46^=(46+4)*(46-4) + 4^2=50*42 + 16=50*(40+2) + 16=2000+100 +16=2100+16=2116
also can be solved as
46^2=50*(50 - 2*4) + 4^2=50*(50-8) +16 = (2500-400) + 16=2100+16=2116
I think that's enough you can try other numbers you wish to, that's easy instead of just cramming squares of numbers till 25 (after you practice and then try to solve it in your mind, you'll find it easy).

By the way yet I didn't add it to the GMAT Quant Notes as the way I explained is obscure and complicated I think so if anyone of you could rephrase it in a better way then I'll add it to the File .
Forgive my English and the way I explained (although I tried my best), haha
Hope it helps! despite the fact I wrote too much to explain a little thing.

Cheers
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Now the above post, how to take square of a number with a little bit correction and squaring of some other 3-digit number, is added to the Quant Notes file. You can download it now.
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This post is related to my post above where I posted the link for "how to check divisibility of a number by 7", where the left to right approach was used. Again, I found right to left approach, which is easier than the right to left approach. Here it is,
Let's take 3528, now first check whether the left most number is divisible by 7 or not. If the left most number is less than 7 then consider nearby number too or you can say now take left most 2-digit number, in this case it is 35.

35/7=5, so we get whole number (0 as a remainder) that means it is divisible by 7, Now remove 35 from the number and replace it with the remainder, in this case it'll become 028 then divide left mos 2-digit number with 7-->02/7=0, we get 0 and the remainder 2, now replace 02 with the remainder 2 and we get 28 now divide 28 with 7-->28/7=4 and we get 4.
So, now collect the answer by placing the bold digits in an order from left to right, and we get 504 as an answer.
Now let's take another number, this time let it be 133.
13/7=1 with 6 as a remainder now replace 13 with the remainder 6 and we have 63 now by dividing 63 with 7-->63/7=9 we get only 9 with no remainder (for right most unit-digit if it is less than 7 then consider right most 2-digits unit & 10th digits as a whole instead of only unit-digit), if there would've been remainder we could say that it is not divisible by 7.
Now the answer will be, after putting bold digits in an order from left right, 19.
so 133/7=19 .
You can try other numbers.
Cheers,
Atif
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