Real Numbers — Exercise 1.1

Method: Divide repeatedly by prime numbers. Stop only when every remaining factor is prime. Group repeated factors into powers.

(i) 140

\[140=2\times70=2\times2\times35=2^2\times5\times7\]

(ii) 156

\[156=2\times78=2\times2\times39=2^2\times3\times13\]

(iii) 3825

\[3825=5\times765=5^2\times153\]

\[153=3\times51=3^2\times17\]

(iv) 5005

\[5005=5\times1001=5\times7\times143\]

(v) 7429

\[7429=17\times437=17\times19\times23\]

For the HCF, take the smallest exponent of every common prime. For the LCM, take the largest exponent of every prime appearing in either number.

(i) 26 and 91

\[26=2\times13,\qquad91=7\times13\]

\[\mathrm{HCF}=13,\qquad\mathrm{LCM}=2\times7\times13=182\]

\[13\times182=2366=26\times91\]

(ii) 510 and 92

\[510=2\times3\times5\times17,\qquad92=2^2\times23\]

\[\mathrm{HCF}=2\]

\[\mathrm{LCM}=2^2\times3\times5\times17\times23=23460\]

\[2\times23460=46920=510\times92\]

(iii) 336 and 54

\[336=2^4\times3\times7,\qquad54=2\times3^3\]

\[\mathrm{HCF}=2\times3=6\]

\[\mathrm{LCM}=2^4\times3^3\times7=3024\]

\[6\times3024=18144=336\times54\]

(i) 12, 15 and 21

\[12=2^2\times3,\quad15=3\times5,\quad21=3\times7\]

Only 3 is common to all three. For the LCM, include every prime at its largest exponent.

\[\mathrm{HCF}=3,\quad\mathrm{LCM}=2^2\times3\times5\times7=420\]

(ii) 17, 23 and 29

These are three distinct primes. They share no prime factor.

\[\mathrm{HCF}=1,\quad\mathrm{LCM}=17\times23\times29=11339\]

(iii) 8, 9 and 25

\[8=2^3,\quad9=3^2,\quad25=5^2\]

There is no prime factor shared by all three.

\[\mathrm{HCF}=1,\quad\mathrm{LCM}=2^3\times3^2\times5^2=1800\]

Check your method: The product rule for HCF and LCM of two integers must not be used unchanged for three integers.

1. For two positive integers, HCF × LCM equals their product.
2. Insert the given HCF and solve for the LCM.

\[9\times\mathrm{LCM}=306\times657=201042\]

\[\mathrm{LCM}=\frac{201042}{9}=22338\]

An independent prime-factor check gives \(306=2\times3^2\times17\) and \(657=3^2\times73\). Taking the highest prime powers gives the same result.

1. Any positive integer ending in zero is divisible by 10, and therefore has 5 as a prime factor.
2. For a positive integer n, the prime factors of 6ⁿ are only 2 and 3.
3. Unique prime factorisation rules out an additional factor 5.

\[6^n=(2\times3)^n=2^n3^n\]

A composite positive integer can be written as a product of two integers greater than 1. Expose a common factor rather than testing many divisors.

(i)

\[7\times11\times13+13=13(7\times11+1)\]

\[=13\times78=1014\]

Both 13 and 78 exceed 1, so this integer is composite.

(ii)

\[7\times6\times5\times4\times3\times2\times1+5\]

\[=5(7\times6\times4\times3\times2\times1+1)\]

\[=5(1008+1)=5\times1009=5045\]

One driver returns to the starting point after 18, 36, 54, … minutes. The other returns after 12, 24, 36, … minutes. The first common return time is the least common multiple.

\[18=2\times3^2,\qquad12=2^2\times3\]

\[\mathrm{LCM}(18,12)=2^2\times3^2=36\]

In 36 minutes, the first completes 2 laps and the second completes 3 laps. Both are back at the starting point.