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NCERT Exemplar · Class 10 Mathematics Real Numbers

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EXERCISE 1.4 1–5 (part 5 of 5)

  1. Exercise 1

    Show that the cube of a positive integer of the form 6q+r,q\displaystyle 6 q+r, q is an integer and r=0,1,2,3,4,5\displaystyle r=0,1,2,3,4,5 is also of the form 6m+r\displaystyle 6 m+r.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[a = 6q+r, \quad r=0,1,2,3,4,5 \] \[a^3=(6q+r)^3=216q^3+108q^2r+18qr^2+r^3 \] \[=6(36q^3+18q^2r+3qr^2)+r^3 \] \[r^3-r=(r-1)r(r+1)=6k \quad \text{(product of 3 consecutive integers)} \] \[a^3=6(36q^3+18q^2r+3qr^2+k)+r=6m+r \] Answer: \(\displaystyle a^3=6m+r\), same remainder \(\displaystyle r\) as \(\displaystyle a\).

  2. Exercise 2

    Prove that one and only one out of n,n+2\displaystyle n, n+2 and n+4\displaystyle n+4 is divisible by 3\displaystyle 3, where n\displaystyle n is any positive integer.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[n=3q,\ 3q+1,\ 3q+2 \quad \text{(Euclid's division lemma)} \] \[n=3q:\quad n=3q,\ \ n+2=3q+2,\ \ n+4=3(q+1)+1 \] \[n=3q+1:\quad n+2=3(q+1),\ \ n=3q+1,\ \ n+4=3(q+1)+2 \] \[n=3q+2:\quad n+4=3(q+2),\ \ n=3q+2,\ \ n+2=3(q+1)+1 \] In each case exactly one term is a multiple of 3. Answer: Exactly one of \(\displaystyle n, n+2, n+4\) is divisible by 3.

  3. Exercise 3

    Prove that one of any three consecutive positive integers must be divisible by 3.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[n=3q,\ 3q+1,\ 3q+2 \quad \text{(Euclid's division lemma)} \] \[n=3q \Rightarrow n=3q \] \[n=3q+1 \Rightarrow n+2=3(q+1) \] \[n=3q+2 \Rightarrow n+1=3(q+1) \] Answer: One of \(\displaystyle n, n+1, n+2\) is always divisible by 3.

  4. Exercise 4

    For any positive integer n\displaystyle n, prove that n3−n\displaystyle n^3-n is divisible by 6.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[n^3-n=n(n-1)(n+1)=(n-1)n(n+1) \] One of any three consecutive integers is divisible by $\displaystyle 3$, and one of any two consecutive integers is divisible by 2. \[\gcd(2,3)=1 \Rightarrow (n-1)n(n+1) \text{ divisible by } 2\times3=6 \] Answer: \(\displaystyle n^3-n\) is divisible by $\displaystyle 6$ for every positive integer \(\displaystyle n\).

  5. Exercise 5

    Show that one and only one out of n,n+4,n+8,n+12\displaystyle n, n+4, n+8, n+12 and n+16\displaystyle n+16 is divisible by 5\displaystyle 5, where n\displaystyle n is any positive integer. [Hint: Any positive integer can be written in the form 5q,5q+1,5q+2,5q+3\displaystyle 5 q, 5 q+1,5 q+2,5 q+3, 5q+4\displaystyle 5 q+4].

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[n = 5q + r, \quad r = 0,\,1,\,2,\,3,\,4 \quad \text{(division algorithm)} \] \[n+4 = 5q + (r+4) \] \[n+8 = 5(q+1) + (r+3) \] \[n+12 = 5(q+2) + (r+2) \] \[n+16 = 5(q+3) + (r+1) \] Remainders of \(\displaystyle n,\ n+4,\ n+8,\ n+12,\ n+16\) on division by $\displaystyle 5$: \[r=0:\ 0,\ 4,\ 3,\ 2,\ 1 \quad (n = 5q) \] \[r=1:\ 1,\ 0,\ 4,\ 3,\ 2 \quad (n+4 = 5(q+1)) \] \[r=2:\ 2,\ 1,\ 0,\ 4,\ 3 \quad (n+8 = 5(q+2)) \] \[r=3:\ 3,\ 2,\ 1,\ 0,\ 4 \quad (n+12 = 5(q+3)) \] \[r=4:\ 4,\ 3,\ 2,\ 1,\ 0 \quad (n+16 = 5(q+4)) \] Each list contains $\displaystyle 0$ exactly once. Answer: Exactly one of \(\displaystyle n, n+4, n+8, n+12, n+16\) is divisible by 5.