Home
Class 10
MATHS
Prove that there is no natural number...

Prove that there is no natural number for which `4^n` ends with the digit zero.

Promotional Banner

Topper's Solved these Questions

  • REAL NUMBERS

    NAGEEN PRAKASHAN ENGLISH|Exercise Exercise 1c|12 Videos
  • REAL NUMBERS

    NAGEEN PRAKASHAN ENGLISH|Exercise Revision Exercise Very Short Answer Questions|12 Videos
  • REAL NUMBERS

    NAGEEN PRAKASHAN ENGLISH|Exercise Exercise1 A|28 Videos
  • QUADRATIC EQUATIONS

    NAGEEN PRAKASHAN ENGLISH|Exercise Revision Exercise Long Answer Questions|6 Videos
  • SOME APPLICATIONS OF TRIGONOMETRY

    NAGEEN PRAKASHAN ENGLISH|Exercise Long Answer Questions|5 Videos

Similar Questions

Explore conceptually related problems

Zero is a natural number.

Consider the numbers 4^n , where n is a natural number. Cheek whether there is any value of n for which 4^n ends with the digit zero.

prove that 2nlt(n+2)! for all natural numbers n.

Prove that ((n^(2))!)/((n!)^(n)) is a natural number for all n in N.

prove that n^(2)lt2^(n) , for all natural number n≥5 .

If (2^(200)-2^(192).31+2^n) is the perfect square of a natural number , then find the sum of digits of 'n' .

The total number of six-digit natural numbers that can be made with the digits 1, 2, 3, 4, if all digits are to appear in the same number at least once is a. 1560 b. 840 c. 1080 d. 480

Number of natural numbers lt2.10^(4) , which can be formed with the digits, 1,2,3 only is equal to

The number of natural numbers n for which (15n^2+8n+6)/n is a natural number is:

The last four digits of the natural number 3^100 are