The number of `4` digit natural numbers such that the product of their digits is `12` is
A. `24`
B. `36`
C. `42`
D. `48`
1 Answers
Talk Doctor Online in Bissoy App
The sum of the digits of a two-digit number is 9. If 27 is subtracted from the number, its digits are interchanged.
Correct option is: C. 18
2 Answers 1 views
A `2n` digit number starts with 2 and all its digits are prime, then the probability that the sum of all 2 consecutive digits of the number is prime i
Correct Answer - B The prime digits are 2, 3, 5, 7. If we fix 2 at first place, then other 2n - 1 places are filled by all four digits....
2 Answers 1 views
A three-digit number is selected at random from the set of all three-digit numbers. The probability that the number selected has all the three digits
Correct Answer - D Three-digit numbers are 100, 101, …, 999. Total number of such numbers is 900. The three-digit numbers (which have all same digits) are 111, 222, 333, …,...
2 Answers 1 views
If `P` is the number of natural numbers whose logarithms to the base 10 have the characteristic `pa n dQ` is the number of natural numbers logarithms
`10^(p) leP lt 10^(p+1)` ` rArr P=10^(p+1) - 10^(p)` ` rArr P=9 xx 10^(p)` Similarly, `10^(q-1)lt Q le10^(q)` `rArr Q = 10^(q)-10^(q-1) = 10^(q-1)(10-1)=9 xx 10^(q-1)` ` :. log_(10)P-log_(10)Q =...
2 Answers 1 views
Find (i) the last digit, (ii) the last two digits, and (iii) the last three digits of `17^(256)dot`
We have `17^(256) = (17^(2))^(128) = (289)^(128) = (290-1)^(128)` `:. 17^(256) = .^(128)C_(0)(290)^(128)-.^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `- .^(128)C_(125)(290)^(3)+.^(128)C_(126)(290)^(2)-.^(128)C_(127)(290)+1` `=[.^(128)C_(0)(290)^(128) - .^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `-.^(128)C_(125)(290)^(3)]+.^(128)C_(126)(290)^(2) -.^(128)C_(127)(290)+1` `=1000m+((128)(127))/(2)(290)^(2)-128xx290+1` `= 1000 m + (128)(127)(290)(145)-128xx290-1` `= 1000m+(128)(290)(127xx145-1)+1` `=1000m+(128)(290)(18414)+1` `=1000m+683527680+1`...
2 Answers 1 views
A binary number is made up to 8 digits. Suppose that the probability if an incorrect digit appearing is `p` and that the errors in different digits ar
Correct Answer - `1-(1-p)^(8)` Probability for an incorrec digit is p. hence, probability for 8 correct digit is `1-p)^(8).` Hence, required probability is `1-(1-p)^(8).`
2 Answers 1 views
Number of four digit positive integers if the product of their digits is divisible by `3` is.
Correct Answer - D `(d)` Product will be divisible by `3` if atleast one digit is , `0,3,6,9` Now total four-digit no.`=9xx10^(3)`, Number of four-digit no. without `0,3,6,9=6^(4)` Total number of...
2 Answers 1 views
The number of four-digit numbers that can be formed by using the digits `1,2,3,4,5,6,7,8` and `9` such that the least digit used is `4`, when repetiti
Correct Answer - B `(b)` Least digit used `=4` `:.` We can use `4,5,6,7,8,9`. So each of the four places can be filled in `6` ways So number of numbers is...
2 Answers 1 views
Number of six-digit numbers such that any digit that appears in the number appears at least twice, where the digits of each number are from the set `{
Correct Answer - C `(c )` Case I : All six digits alike i.e. `111111`, `222222`……..etc. `=5` ways Case II : `2` alike `+2` other alike. Select any three in `"^(5)C_(3)`...
2 Answers 4 views
The number of `n` digit number formed by using digits `{1,2,3}` such that if `1` appears, it appears even number of times, is
Correct Answer - B `(b)` Required number of numbers `="(n)C_(0)(2)^(n)+"(n)C_(2)(2)^(n-2)+…` We know that `(1+x)^(n)+(x-1)^(n)=2sum_(k=0)^(n)"(n)C_(k)x^(k)` Hence, put `x=2`, get the result `2["^(n)C_(0)(2)^(n)+^(n)C_(2)(2)^(n-2)+^(n)C_(2)(2)^(n-4)+^(n)C_(6)(2)^(n-6)+...]`
2 Answers 1 views