Every Number Sense test is split into roughly four sections of 20 questions, and each of these
4 sections generally has it's own certain problem types that will always appear (specifically: https://docs.google.com/document/d/12wzaJ2WjTT-gLsnaX1pJLGT_CYo1g82D604WIzinQqA/edit?usp=sharing). So, I've split up the tricks in a similar fashion. Focus the most on the first two sections as perfecting these will get you far enough to make region.
Speed is the most important factor in this section, because a lot of the problems are just normal arithmetic. The most important thing to practice for this is the memorization as well as the multiplication tricks.
1. Adding a/b + b/a : 2 +( (a-b)^2)/ab. Eg 5/8 + 8/5 = 2 + (9/40)
2. Subtracting reverse numbers : 3 digit - Subtract the last digit from first, multiply by 100 and then subtract the same difference. Eg. 672 - 276 = (4*100) - 4 = 396 ( For 4 digit number - same trick multiply by 1000)
3. Mixed number subtraction : If the fraction part of 1st number is larger then the whole number and fraction parts can be subracted seperately. Eg. 8 3/5 - 5 3/8 = ( 8-5) ( 3/5-3/8) = 3 9/40
4. When adding several numbers with no obvious pattern : try to find pairs which add up easily Eg. 34 + 51 + 68 + 12 + 29 + 46 = (34+46) + (51+29) + (68+12) = 80 + 80 + 80 = 240
1. If tens digit is the same and units digits add to 10 (Eg xy * xz), y+z = 10 : the product of the tens digits is the last 2 digits ( y*z) and (x)*(x+1) is the first 2 digits. Eg 74*76 = (7*8)(4*6) = 5624.
2. If units digit is the same and tens digits add to 10. Product of units digits are last 2 digits. First 2 digits = product of tens digits + units digit Eg 67 * 47 = (( 6*4) + 7)(7*7) = 3149
3. Multiply by 25 (*) : Divide by 4 and multiply by 100 ( for 50, half and multiply by 100). Similary to multiply by 2.5 and 5, multiply by 10 and divide by 4 or 2. To multiply by 75, get 3/4 of the number and multiply by 100.
4. Multiply by 101 : xy * 101 = xyxy
5. Multiply by 11 (*): Two ways to do this:
a. Add a zero after the number and then add the number. Eg 56* 11 = 560 + 56 = 616
b. xyz * 11 = x(x+y)(y+z)z Eg. 358*11 = 3(3+5)(5+8)8 = 3938. Since we have to add the carrover, always go from right to left.
6. Multiplying numbers close to 100 :
a. Both > 100 : product of units dgits = last 2 digits of result, any one of the original numbers + units digit of other = first 3 digits
b. Both < 100 : multiply the difference of each term from 100 = last 2 digits. 100 - Sum of differences = first 2 digits
c. One > and 1 < 100 : multiply difference of each term from 100. 100 - minus this product = last 2 digits. First 2 digits = larger number - ( difference of smaller from 100) - 1
7. a * a/b = Whole number part = ( a ) + ( a-b), Fraction part = (a-b)^2/b
8. Multiplying numbers that end in 5 : Sum the tens digits, if they add to even then the last 2 digits are 25, else 75. First 2 digits are ( product of tens digits) + ( average of tens digits)
Eg : 55 * 65 : Sum of tens is 11, so last 2 digits = 75, first 2 digits = ( 5*6) + (11/2) = 30 + 5 ( if average has .5, then round down ) = 35. Answer = 3575
9. Multiplying Reverses : This trick works for 2 double digit numbers - ab*ba. The units digits is the 2 digits multiplied, tens digit is the sum of the squares of the digits and the hundreds digit is the 2 digits multiplied. ( carry over as needed)
Eg. 67 * 76 : Units digit is 42, (Write down 2 carry 4), tens is (36 + 49) = 85 + 4 = 89 (Write down 9 carry 8), hundreds is 42 + 8 = 50. Answer is 5092
10. Multiplying two numbers equidistant from a third: You can use difference of two squares here.
Eg. 17 * 23 = (20-3)(20+3) = 20^2 - 3^2 = 391
1. Squaring any 2 digit number ab (*): last digit is b^2, middle digit is 2ab, first digit is a^2 ( add carryover wherever applicable)
Eg. 72 - last digit is 4, middle is 2*2*7 = 2(8), first is 49+2 = 51 ( 5184).
This trick also works for a 3 digit number with 0 as middle number:
Eg. 304 - last 2 digits are 16 , middle is 2*3*4 = 24 first is 9 (92416)
2. Squaring numbers ending in 5 (*) eg a5 - last 2 digits are 25, first 2 digits - a*(a+1)
Eg. 35^2 is (3*4)25 = 1225.
The same trick applies to muliplying ANY 2 numbers whose tens digit is the same and units digit add to 10.
Eg. 64 * 66 = (6*7) 6*4 -> 4224
3. For 12,21 and 13 ,31 the squares are reverse - 12 = 144, 21 = 441 and 13 = 169, 31 = 961
1. When one number is twice the other (*) - (smaller number ^ 2 ) * 5
Eg. 12^2 + 24^2 = 144*5 = 720
2. When one number is thrice the other (*) - (smaller number ^ 2 ) *10
Eg. 12^2 + 36^2 = 144*10 = 1440
3. 2 consecutive numbers x,y : 2*x*y + 1
Eg. 14^2 + 15^2 = 2*14*15 + 1 = 420 +1 = 421
Another way to do this ( esp if the numbers are big) : 2* x^2 + 2*x + 1
4. When the 2 numbers are equidistant from a third number: Eg 17^2 + 23^2 : both numbers are equidistant from 20, and the difference is 3: The answer is 2(20^2 + 3^2 )= 2(409 ) = 818
5. When the 2 numbers follow the rule : Inner 2 digits are 1 apart and outer 2 digits add to 10 (*) : Take the number in which the inner digit is larger , add the squares of the individual digts of this number and then multiply the result by 101. Eg. 94^2 + 31^2 : 4 is the larger inner digit, so take 94, 101( 9^2 + 4^2) = 101 ( 97) = 9797
6. Difference of two squares: a^2 - b^2 = (a+b)(a-b)
Eg. 54^2 - 55^2 = (54-55)(54+55) = (-1)(109) = -109
1. Divide by 3 or 9 : Add the digits of the number until you get a number less than 3 or 9. That will be the remainder.
Eg. 456 % 9 -> 4 + 5 + 6 = 15 -> 1 + 5 = 6, so 456 = 6 (mod 9)
2. Divide by 4 or 8 : there is a remainder when the last 2 digits of the number are divisible by 4 or the last 3 digits of the number are divisible by 8
Eg. 1800 is divisible by 8 because 800 is divisible by 8.
3. Divide by 6 : Use the remainder rules for 2 and 3, if the remainder is 0 for both then the number is divisible by 6. Otherwise you can use the results to see what the remainder will be.
Eg. 556677 ÷ 6 -> diving by 2, there is a remainder of 1, and dividing by 3, there is a remainder of 0. So, the remainder that is odd but divisible by 3 will be 3.
4. Divide by 11 : You add up alternating digits (beginning with the ones digits) then subtract the sum of the remaining digits
Eg. 13542 ÷ 11 -> (2+ 5 + 1) = 8, (3+4) = 7, 8-7 = 1
5. For problems to find remainder of an expression (Eg. 23+67*45 (mod 3) ), find the remainder of individual numbers and then do the calculation. In this case: (2 + 1*0) = 2
1. LCM (x,y) * GCD (x,y) = x*y
2. GCD (x, y) - Divide larger by smaller and get the remainder, then keep dividing the smaller by remainder, till the remainder is 0.
Eg. GCD ( 54, 90 ) = 90/54 - remainder is 36, 54/36 - remainder is 18, 36/18 ( 0 remainder) - so 18 is the GCD
1. .ababab = ab/99
2. .abcabc = abc/999
3. .abbb = (ab-a)/90
4. .abcbcbc = (abc - a)/990
Eg. If 14/33=.ababab, Find a+b: 14/33 = 42/99 , so a =4 and b = 2, a+b = 6
1. Prime Numbers - 2,3,5,7,11,13, 17, 19,23, 29, 31, 37, 41,43,47,53,59,61,67,71,73,79,83,89,97 ( For questions like how many primes are there less than 30 etc- memorize how many there are in each group of 10, ( 4,4,2,2,3,2,2,3,2,1)
2. Fibonnaci - 1,1,2,3, 5, 8, 13, 21, 34, 55, 89
a. sum of first 10 terms in any fibonnaci sequence = 7th term * 11
b. sum of first n fibonnaci = (n+2) - second term
c. sum of squares of 1st n fibonnaci = n * (n +1)
3. For sums of equally spaced numbers, multiply the median of the numbers by the number of terms. Eg. 24+27+30+33+36 = 30*5 = 150
4. Sum of terms in arithmetic sequence - Let d be the common difference, a_1 be the first given term, and a_n be the last given term. The sum is ((a_n - a_1)/d + 1) * (a_1 + a_n) / 2- This is essentially just the number of terms times the median
Additive inverse of x = -x, Multiplicative Inverse of x = 1/x
1.To covert a number from base 10 to any other base, just keep dividing the number by the other base and put the remainder down, till you reach 0.
Eg. 118 base 10 to base 6 : 118/6 = 19 (4), 19/6 = 3(1), 3/6 = 0(3) - answer is 314 base 6
2. Base 2 to base 8 : Since 2^3 = 8, divide the number into groups of 3 and then convert to base 10 and keep putting down the result.
Eg. 111001 base 2 to base 8 : split it into 111 and 001, convert each to base 2, which is 7 and 1 respectively. So answer is 71 base 8
3. Base 3 to base 9 : 3^2 = 9, divide into groups of 2 and convert to base 10.
Eg. 12210 base 3 to base 9, 01,22, 10 converting each to base 10, we get 1, 8, 3. Answer is 183
1. To find the number of positive integral divisors of any number : Prime factorize, take each exponent +1, and then muliply them all.
Eg. For 45 - Prime factorization is 3^2 * 5^1 - exponents are 2, 1. Multiply (2+1)(1+1) = 6
2. Sum/Product of Coefficients of a polynomial :
a. Sum of coeffs : Replace all variables by 1 and solve. Eg. (2x - 3y)^5 = (2-3) ^5 = (-1)^5 = -1
b. Product of coeffs : Use the memorized binomial coeffs. and solve. Eg : (2x+y)^3 - Here the binomial coeffs are 1,3,3,1. So it solves to (2^3 * 1) ( 2^2 * 3) (2*3)(1)= 8*12*6*1 = 48*12 = 576
3. Sum/Product of Roots :
a. Sum of roots: -b/a Eg. Sum of roots of 3x^2 - 2x + 5 -> -(-2) / 3 = 2/3
b. Product of roots: c/a for 2nd degree and -d/a for third degree. Eg. Product of roots of 2x^3 - 3x^2 + 4x - 6 = -(-6) / 2 = 3
4. Solving Equations : Here the trick mostly is not to directly substitute the given value, because it will be a simple factorization which will make it quicker.
Eg. If f(x) = 9x^2 - 12x + 4 then f(8) = ?. This is a simple factorization (3x-2)^2 = (24-2)^2 = (22)^2 = 484
If f(x) = x^3 + 3x^2 + 3x +1 then f(11) = ?. This is clearly (x+1)^3 = (12)^3 = 1728
5. Estimating Square/Cube Roots :
a) For square roots, you can take out groups of two digits and approximate them as two zero's until you have 1/2 digits left.
Eg. Approximate the square root of 74125 -> You can approximate the 25 and 41 as two sets of 0's, which gives you 100 out of the square root. Then, the square root of 7 is approximately 2.6, so we get 100 * 2.6 = 260.
b) For cube roots, you take out groups of 3 digits instead of 2.
Eg. Approximate the cube root of 60130224. We take out 224 and 130 and approximate them as groups of 0's, giving us 100 out of the cube root. Then, the cube root of 60 is slightly less than 4, so we get 390.
1. Geometric Mean- nth root of the product of the n terms
2. Harmonic Mean- n/sum of reciprocals of terms.
3. Sum of a number and its square- a + a^2 = a*(a+1)
Eg. 49^2 + 49 = 49*50 = 2450, Similarly a^2 - a = a*(a-1)
4. Decimal to fraction:
a. abc/999 = .abcabc
b. abc/900 : first digit is a, second is (a+b), 3rd is (a+b+c) and then this is the repeating number
c. abc/990 : ab are the first 2 digits, third is (a+c), now the 2nd and 3rd digits are the repeating number
5. Infinite Geometric Sequences:
a. If each term is 2/3 of the previous term - the sum is just 3* first term
b. If each term is 1/2 the previous term - the sum is just 2* first term
c. For any other one, the sum is (first term) / (1 - common ratio)
These are obviously not all the tricks you'll have to know, just the ones I think show up most commonly and are most important. Perfection of the memorization section and these tricks will get you REALLY far. For anyone who wants to go above and beyond, check the additional section for other resources.
CWHS Number Sense (Sam Singh)