Tricks for faster calculations - Squares of 2 digit numbers

Let ab be the number. i.e. Digit at Ten's place be a and digit at Unit's place be b.
1) Calculate square of a. Put 2 dots(..) to it's right.
2) Calculate 2 x a x b. Put 1 dot and write below above number such that right most dots in both steps are one below the other.
3) Now calculate square of b and write its unit place's digit below the right most dot written in step 2). Write rest of number below units place digit of number written in step 2)
4) Make simple addition. Treat dots as ZEROS.

Ex. 89^2
1) 8 x 8 =64 ------------------- 6 4  .  .
2) 2 x 8 x 9 = 144-------------- 1 4 4  .
3) 9 x 9 = 81 ------------------        8 1
-------------------------------------------
                                                  7 9 2 1
4) Add to get required square.

New method for division

NEW METHOD OF DIVISION.

Let us learn first how to find divisibility test for a prime number? This will lead us to the new method of division.

Divide 100 , 1000 , 10000 by the prime number. Note quotients and remainders.

Select that set of quotient and remainder where the remainder and quotients are both small numbers. It should be preferably single digit.

Suppose we want to find divisibility test for 13.

(a) 100/13 = 7; remainder 9. 13 x 7 + 9 = 100.

(b) 1000/13=76; remainder 12 or 13 x 76 +12 = 1000

(c) 10000/13=769; remainder 3

Best selection will be 7 & 9. i.e. for 100 since we are getting pair of small numbers that will be easy for calculation.

Now, any natural number (dividend) can be expressed as: 1000a+100b+10c+d , where a,b,c,d are from 0 to 9.( Whole number up to 9.i.e. 0 <= a, b, c ,d <=9 )

Now 1000a+100b+10c+d = 910a+90a+91b+9b+10c+d.

(1) Since we selected the pair of quotient and remainder for 100; leave aside last 2 digits(= no. of zeros to the right of 1 in 100).

(2) 90a+9b=9(10a+b). So multiply the number left after leaving aside last 2 digits,  by 9 (remainder).

(3) Add last 2 digits [from step (1)] to this multiplication.

(4) Divide the sum by 13. Note Quotient and Remainder. If remainder is zero the number (dividend) under consideration is divisible by 13.

(5) We know 910a+91b will be divisible by 13.and answer of division will be (a) Quotient= 7x(10a+b) i.e. 7 x the number left after leaving aside last 2 digits; (b) remainder zero since 91=13 x 7. 7 is fixed part of test.

(6) Add the quotients in step (4) & (5). This is the quotient when the dividend is divided by 13.

(7) When the number is divided by 13, the remainder will be same as that obtained in step (4).

(8) Note the test can be used as another method for division. In this method,

the only  advantage is that dividend has 2 digits less than the original number.

Engineered by

Hemant Shashikumar Dikshit

Geometry rider

If in triangle ABC, seg AB is congruent to seg AC and AP , BQ & CR are angle bisectors of angles A, B & C respectlively; (NOTE = ( is congruent to) )
prove that 1) seg CQ = seg BR
2) seg AR = seg AQ
3) seg RQ is parallel to side BC
4) Triangle ARQ is similar to triangle ABC
5) Triangle RBC is congruent to triangle QCB
6) Triangle PQR is an isosceles triangle.