NCERT Exemplar Class 10 Maths Chapter 1 Real Numbers Exercise 1.4 Solutions

Binomial then a residue check. The proof splits neatly into one expansion and one small table.

• Expand: expanding (6q+r)³ produces three terms each divisible by 6 plus the lone r³, so the cube has the form 6k+r³.
• Check the leftover: the only question left is whether r³ keeps the same remainder r modulo 6, and a direct table for r=0 to 5 returns the remainders 0,1,2,3,4,5, identical to the inputs.
• Closure: so the cube carries the very same form 6m+r as the original number, which is a clean closure property of cubing modulo 6.

(6q+r)³=6m+r for every r=0,,5.

They cover all residues mod 3. The slick argument looks at the three numbers as residues rather than as values.

• Reduce mod 3: modulo 3 the three numbers become n, n+2 and n+1, because 4 is the same as 1, so they form the three distinct residues 0,1,2 in some order.
• Exactly one zero: since each residue appears once, exactly one of them is 0 modulo 3, and that single zero residue marks the one multiple of 3.
• Why only one: there can be no second multiple, which is what proves ``one and only one'', and the case table above is just this same idea spelled out longhand.

Precisely one of the three is a multiple of 3.

A complete residue run. Consecutive numbers automatically sweep through every remainder mod 3.

• Sweep the residues: three numbers in a row step through the remainders r, r+1 and r+2 modulo 3, which is the full set 0,1,2.
• One is zero: the residue 0 shows up exactly once, marking the multiple of 3, so one of n, n+1, n+2 is divisible by 3 no matter where the block starts.
• Why it matters: this single fact is the backbone of the result that a product of three consecutive integers is divisible by 6.

One of three consecutive integers is a multiple of 3.

One factorisation, two guarantees. A single clever factorisation turns a cubic into three numbers in a row.

• Factor it: writing n³-n as (n-1)n(n+1) exposes three consecutive integers sitting side by side.
• Two factors appear: such a triple always contains a multiple of 2 and a multiple of 3, exactly the facts proved in Q13 and Q37, so both small factors are present.
• Combine: because 2 and 3 are coprime, their product 6 must divide the whole expression, so 6 divides n³-n for every positive integer n with no exceptions left to check.

6n³-n for all n.

Five offsets, five residues. The five offsets are chosen so they sweep every remainder mod 5 exactly once.

• Reduce mod 5: reducing the five numbers n, n+4, n+8, n+12, n+16 modulo 5 gives r, r+4, r+3, r+2, r+1, which is the complete residue set 0,1,2,3,4 in some order.
• One hits zero: exactly one of these residues equals 0 modulo 5, so exactly one of the five numbers is divisible by 5 and never two.
• Table agrees: the five-case table above just confirms this offset by offset, with the divisible member shifting as the starting residue r changes.

One and only one of the five is divisible by 5.