# Maths

**Multiplication by 5
**It's often more convenient instead of multiplying by 5 to multiply first by 10 and then divide by 2.

For example, 137·5=1370/2=685.

**Division by 5
**Similarly, it's often more convenient instead to multiply first by 2 and then divide by 10.

For example, 1375/5=2750/10=275.

**Division/multiplication by 4
**Replace either with a repeated operation by 2.

For example, 124/4=62/2=31. Also, 124·4=248·2=496.

**Division/multiplication by 25
**Use operations with 4 instead.

For example, 37·25=3700/4=1850/2=925.

**Division/multiplication by 8
**Replace either with a repeated operation by 2.

For example, 124·8=248·4=496·2=992.

**Division/multiplication by 125
**Use operations with 8 instead.

For example, 37·125=37000/8=18500/4=9250/2=4625.

**Squaring two digit numbers.
**You should memorize the first 25 squares:

**Squares of numbers from 26 through 50.
**Let A be such a number. Subtract 25 from A to get x. Subtract x from 25 to get, say, a. Then A

^{2}=a

^{2}+100x.

For example, if A=26, then x=1 and a=24. Hence 262=242+100=676. Similarly, if A=37, then x=37-25=12, and a=25-12=13. Therefore, 372=132+100·12=1200+169=1369. Why does this work? (25+x)2-(25-x)2=·=50·2x=100x.

**Squares of numbers from 51 through 99.
**The idea is the same as above. (50+x)2-(50-x)2=100·2x=200x.

For example, 632=372+200·13 = 1369+2600 = 3969.

**Squares of numbers from 51 through 99, second approach**.

We are looking to compute A2, where A=50+a. Instead compute 100·(25+a) and add a2. Example: 572. a=57-50=7. 25+7=32. Append 49=72. Answer: 572=3249.

In general, a2 = (a + b)(a - b) + b2. Let a be 57 and, again, we wish to compute 572. Let b = 3. Then 572 = (57 + 3)(57 - 3) + 32, or 572 = 60·54 + 9 = 3240 + 9 = 3249.

**Squares of numbers that end with 5.
**Let A=10a+5. Then A2=(10a+5)2=100a2+2·10a·5+25=100a(a+1)+25.

For example, to compute 1152, where a=11, first compute 11·(11+1)=11·12=132 (since 3=1+2). Next, append 25 to the right of 132 to get 13225! Another example, to compute 2452, let a=24. Then 24·(24+1)=242+24=576+24=600. Therefore 2452=60025. Here is another way to compute 24·25: 24·25=2400/4=1200/2=600. The rule naturally applies to 2-digit numbers as well. 752=5625 (since 7·8=56).

**Product of two one-digit numbers greater than 5.
**This is a rule that helps remember a big part of the multiplication table. Assume you forgot the product 7·9. Do this. First find the access of each of the multiples over 5: it's 2 for 7 (7 - 5 = 2) and 4 for 9 (9 - 5 = 4). Add them up to get 6 = 2 + 4. Now find the complements of these two numbers to 5: it's 3 for 2 (5 - 2 = 3) and 1 for 4 (5 - 4 = 1). Remember their product 3 = 3·1. Lastly, combine thus obtained two numbers (6 and 3) as 63 = 6·10 + 3.

The explanation comes from the following formula:

(5 + a)(5 + b) = 10(a + b) + (5 - a)(5 - b)

In our example, a = 2 and b = 4.

**Product of two 2-digit numbers.**

If the numbers are not too far apart, and their difference is even, one might use the well known formula (a+n)(a-n)=a2-n2. a here is the average of the two numbers.

For example, 28·24=262-22=676-4=672 since 26=(24+28)/2. Also, 19·31=252-62=625-36=589 since 25=(19+31)/2.

If the difference is odd use either n(m+1)=nm+n or n(m-1)=nm-n.

For example, 7·34=37·35-37=362-12-37=1296-1-37=1258. On the other hand, 37·34=37·33+37=352-22+37=1225-4+37=1258.

**Product of numbers that only differ in units**.

If the numbers only differ in units and the sum of the units is 10, like with 53 and 57 or 122 and 128, then think of them as, say 10a+b and 10a+c, where b+c=10. The product (10a+b)(10a+c) is given by 100a2+10a(b+c)+bc =&nbs;100a(a+1)+bc. Thus to compute 53 times 57 (a=5, b=3, c=7), multiply 5 times (5+1) to get 30. Append to the result (30) the product of the units (3·7=21) to obtain 3021. Similarly 122·128 = 12·13·100+2·8=15616.

**Multiplying by 11.
**To multiply a 2-digit number by 11, take the sum of its digits. If it's a single digit number, just write it between the two digits. If the sum is 10 or more, do not forget to carry 1 over.

For example, 34·11=374 since 3+4=7. 47·11=517 since 4+7=11.

**Faster subtraction**.

Subtraction is often faster in two steps instead of one.

For example, 427-38=(427-27)-(38-27)=400-11=389. A generic advice might be given as "First remove what's easy, next whatever remains". Another example: 1049-187=1000-(187-49)=900-38=862.

**Faster addition**.

Addition is often faster in two steps instead of one.

For example, 487+38=(487+13)+(38-13)=500+25=525. A generic advice might be given as "First add what's easy, next whatever remains". Another example: 1049+187=1100+(187-51)=1200+36=1236.

**Faster addition, #2.
**It's often faster to add a digit at a time starting with higher digits.

For example, 583+645=583+600+40+5=1183+40+5=1223+5=1228.

**Multipliply, then subtract**.

When multiplying by 9, multiply by 10 instead, and then subtract the other number.

For example, 23·9=230-23=207. The same applies to other numbers near those for which multiplication is simplified. 23·51=23·50+23=2300/2+23=1150+23=1173. 87·48=87·50-87·2=8700/2-160-14=4350-160-14=4190-14=4176.