Extended Concepts (Unit Digit-GMAT)

Extended Concepts (Unit Digit-GMAT)

The units digit of a number is the last digit of the number before the decimal point. For example, 9 is the units digit of 79, and 6 is the units digit of 76546. there is an interesting rule for multiplication and division for unit digits.

its uses for

i. when we select answer choice after long calculations.

ii. finding the remainder when it divided by 5 or 10.

Example: The last digit of 85945×89×58307=5×9×7=45×7=35=5?

Interesting rules for Unit digits.

Example: The last digit of 85945×89×58307=5×9×7=45×7=35=5?

Cyclisity of Numbers

One Cyclisity                0                       1                 5                   6

 Two Cyclisity              41[odd]   = 4             91[odd]   = 9                                         

                                    42[even]   = 6             92[even]    = 1

Four  Cyclisity          

              21 =    2           31 =   3         71 =    7          81= 8

              22 =    4           32 =   9          72 =    9         82 = 4

              23 =    8           33 =   7          73 =     3        83 = 2

              24 =     6          34 =   1          74 =     1        84 =  6

If exponent is large then divide it by 4 and put remainder as exponent. If remainder is zero then put 4 as exponent.

[i] whatever exponent of number having cyclicity 1

Unite digit = number itself

[ii] odd exponent of number having 2 cyclicity,

unit digit = number itself

[iii] even exponent of number having 2 cyclicity,

unit digit = exponent 2

[iv] for number having cyclicity 4, first divide exponent by 4 then,

Reminder 1 = expo 1

Reminder 2 = expo 2

Reminder 3 = expo 3

Reminder 0 = expo 4

To find the number of prime or number of zero 

It’s easier if you look at an example:

How many zeros are in the end (after which no other digits follow) of  32!?

32/5 +32/52 = 6+1 =7 (denominator must be less than 32, 52 < n is less) Hence, there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n! but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

Finding the number of powers of a prime number p, in the n!

The formula is: n/p +n/p2 +n/p3 +…… till px < n

What is the power of 2 in 25!?

25/2 +25/4 +25/8 +25/16 =12 + 6 + 3 + 1 =22

Finding the power of non-prime in n!:

How many powers of 900 are in 50!

Make the prime factorization of the number: 900 =22×32×52 , then find the powers of these prime numbers in the n!.

Find the power of 2:  50/2 +50/4 +50/ 8 +50/32 =25+ 12+ 6+ 3+ 1=47

Find the power of 3: 50/3 +50/9 +50/27 = 16+ 5 +1=22

Find the power of 5: 50/5 +50/25 = 10 +2 =12

We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900

in the power of 6 in 50!.

Examine Manually and using formula 17! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 × 12 × 13 × 14 × 15 × 16 × 17 Find number of 2 = 15 manually By Formula = 17/2 + 17/4 + 17/8 + 17/16 = 8 + 4 + 2 + 1 = 15 Number of 3 = 17/3 + 17/9 = 5 + 1 = 6 Number of 5 = 17 /5 = 3 No. of 2 > no. of 3 > no. of 5 It is only applied for prime number. For non-prime numbers find the largest prime factor only. If we have to find a number of 15= 5 × 3 then we only need to find 5 If we have to find a number of 10= 5 × 2 then we only need to find 5 If we have to find a number of 8 = 2 × 2 × 2 then we find 2 and divide it by 3

 
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