This is a complementary page where we endeavour to provide some mathematics and problem solving tips and tricks. The questions deal with standard aptitude test describing methods such as efficiency method, total effort method, mathematical induction and so on.
Divisibility by 2: A number is divisible by 2 if digit at unit place is even.
Divisibility by 3: A number is divisible by 3 if sum of all digits is also divisible by 3. E.g. 1323 is divisible by 3 because (1 + 3 + 2 + 3) = 9 which is divisible by 3.
Divisibility by 4: A number is divisible by 4 if number comprising of digits at tens and units is divisible by 4. E.g. 123456 is divisible by 4 because 56 is divisible by 4.
Divisibility by 5: A number is divisible by 5 if digit at unit place is either 0 or 5.
Divisibility by 6: A number is divisible by 6 (2 x 3) if [a] digit at unit place is an even number - divisibility by 2 and [b] sum of all digits is divisible by 3 - divisibility by 3.
Divisibility by 8: A number is divisible by 8 if number comprising of digits at hundreds, tens and units is divisible by 4. E.g. 123456 is divisible by 8 because 456 is divisible by 8.
Divisibility by 9: A number is divisible by 9 if sum of all digits is also divisible by 9. E.g. 1323 is divisible by 9 because (1 + 3 + 2 + 3) = 9 which is divisible by 9.
The numerator is repeated, decimal place = number of digits less by 1. Example:
Step-1: 513/9 = 513 513 513 513 repeat the numerator. Step-2: Location of decimal place = 2 (no of digits in 513) - 1. Thus, 513/9 = 51.3513
Divisibility by 11: A number is divisible by the difference in sum of digits at alternate place is divisible by 11. E.g. 7238561 is divisible by 11 because [1+5+3+7] - [6+8+2] = 0 which is divisible by 11.
Divisibility by 12: A number is divisible by 12 (3 x 4) if [a] sum of all digits is also divisible by 3 - divisibility by 3 and [b] number comprising of digits at tens and units is divisible by - divisibility by 4.
Number of digits in products
Product of integers with N and M digits
N = 2, M = 3: 10 * 100 = 1000, D = 4 = [N + M - 1]
N = 2, M = 3: 99 * 100 = 9900, D = 4 = [N + M - 1]
N = 2, M = 3: 10 * 999 = 9990, D = 4 = [N + M - 1]
N = 2, M = 3: 99 * 999 = 98901, D = 5 = [N + M]
N = 2, M = 4: 10 * 1000 = 10000, D = 5 = [N + M - 1]
N = 2, M = 4: 99 * 1000 = 99000, D = 5 = [N + M - 1]
N = 2, M = 4: 10 * 9999 = 99990, D = 5 = [N + M - 1]
N = 2, M = 4: 99 * 9999 = 989901, D = 6 = [N + M]
Thus: the answer is [N + M - 1] and [N + M].
Square of a number
a2 = (a + b)*(a - b) + b2. For example:
43 * 43 = (43+3)*(43-3) + (3*3) = (46*40)+9 = 1849.
A great improvisation of this formula as per Vedic mathematics is as follows: "Whatever the extent of its difficiency, lessen it still further to that very extent; and also set-up the square of that deficiency". The calculations steps are then defined as:
Take the power of (10, 100, 1000 ...) nearest to the number being squared. E.g. n = 84 has base number b = 100.
Calculate deficiency, d = b - n = 100 - 84 = 16
Calculate square of the deficiency: d2. Write the right two digits, xx and carry the third digit if any say c. Here, d2 → 162 → xx = 56, c = 2
Subtract twice of the defficienty from base and add the carry c. Thus: aa = b - 2 * d + c. Here: a = 100 - 2 * 16 + 2 = 70.
Thus: 842 = aa|xx = 70|56 = 7056
Note that this method is best suitable if the diffciency from suitable base (10, 100, 1000 ...) is in single digit. Some adapatation will be presented later.
Followings are the specific applications of this rule:
Square of any number ending is 5: eg. x52 = x*(x+1)|25. Thus: 752 = 7*8|25 = 5625. 1152 = 11*12|25 = 13225
Square of numbers, when the sum of digits at unit's place is 10 and other digits are same, can be calculated using similar method. That is xy * xq where y + q = 10. xy * xq = x * (x+1)|p*q. For example: 67 * 63 = 6*(6+1)|7*3 = 4221.
Note: 24 raised to an even power always ends with 76, 24 raised to an odd power always ends with 24, 76 raised to any power ends with 76.
Large Exponent of a Large Number
10012 = 1002001. Hence, 10012 - 1 is divisible by 1000 or multiple of 1000 as per divisibility rules of 2, 3, 4 ...
10013 = 1003003001. Hence, 10013 - 1 is divisible by 1000 or multiple of 1000 as per divisibility rules of 2, 3, 4 ...
Approximate Square Root of a number
Last digit of power of 3, 4, 7 ...
34 = 81
35 = 243
36 = 729
37 = 2187
38 = 6561
3n = 1 if remainder(n/4) = 0
= 3 if remainder(n/4) = 1
= 9 if remainder(n/4) = 2
= 7 if remainder(n/4) = 3
Similarly, last digit of 4n is 6 is n is an even number and 4 if n is an odd number.
Similarly, last digit of 7n
7n = 1 if remainder(n/4) = 0
= 7 if remainder(n/4) = 1
= 9 if remainder(n/4) = 2
= 3 if remainder(n/4) = 3
Find tens place of any exponent of a number ending in 1
E.g. tens place of 3157. The unit place would be 1. The tens place would be 3 * 7 = 21 and hence 1 will be at tens place. The number is calculated at product of digit at tens place in the base that is '3' here and the first digit (at units place) in the exponent which is '7' here.
So far easy for exponents. How to find digit at tens place in multiplication of 3 or 4 digits? Refer to the method below for product 24 × 37 × 68 × 94.
Work and time - Use efficiency as short-cut
If A does the task in 20 days and B does it in 16 days. How long will they take it to complete the task together?
Another variant of such problem is given as: If a group-A of 4 persons each of equal efficiency can do a task in 12 days. How many day would be required to do the same task by group-B of 6 persons with efficiency 2 times that of group-A?
Equality of total effort Let x be the efficiency of persons in group A. Total effort required = 4 * x * 12 = 48x [mandays]. This is constant for that particular task.
Now, let N be the number of days required for persons in group-B to complete the task. Thus: N * 2x * 6 = 48x. Hence, N = 4 days.
Sum of Arithmetic progression of numbers and sqaures
Sum of A.P. of squares
Sum of A.P. of Cubes and Products
There are two containers having equal volume of A and B. Now, amount x of liquid A from container is taken out and mixed with liquid B. Thereafter, same volume x is taken out from second container and mixed with liquid A. Question is: which of the containers have higher impurity? In other words, is the volume fraction of B in A in the first container is higher, lower or same as volume fraction of A in B in the second container? Answer: The volume fraction of B in A in container 1 = volume fraction of A in B in container 2. Before the answer is demonstrated through complex calculation, the easiest way to get to the answer is by noticing the variable 'x'. It can be between 0 to 100. Hence, assume the case when x = 100 and the answer is straightforward.
Now mathematics is as follows:
What is the shortest path between two diagonally opposite vertices of a cube when the travel path has to be all along the walls, ground and ceiling of the cube? Two such set of vertices have been shown by circles and diamonds in the following figure.
Other than the trivial examples of 0 and 1, the only natural numbers that equal the sum of the cubes of their digits are 153, 370, 371, and 407.
Kaprekar’s Constant - Take any four-digit number except an integral multiple of 1111 (i.e., don’t take one of the nine numbers with four identical digits). Rearrange the digits of your number to form the largest and smallest strings possible. That is, write down the largest permutation of the number, the smallest permutation (allowing initial zeros as digits), and subtract. Apply this same process to the difference just obtained. Within the total of seven steps, you always reach 6174. At that point, further iteration with 6174 is pointless: 7641–1467 = 6174. Example: Start with 8028. The largest permutation is 8820, the smallest is 0288, and the difference is 8532. Repeat with 8532 to calculate 8532–2358 = 6174. Your own example may take more steps, but you will always reach 6174.
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