How to use this set

Here are twelve ques­tions cov­er­ing the whole chap­ter: fac­tori­sa­tion, HCF and LCM, and irra­tional­ity. Try each one hon­estly before you look at its answer. When a ques­tion says "with­out divid­ing" or "with­out fac­toris­ing", take that seri­ously; the restric­tion is the whole point. It checks whether you can rea­son from prime fac­tori­sa­tions instead of grind­ing through the arith­metic.

One piece of advice from expe­ri­ence: in the exam, marks are lost far more often to a wrong prime fac­tori­sa­tion than to a wrong method. So check every fac­tori­sa­tion by mul­ti­ply­ing back.

Sec­tion A, Fac­tori­sa­tion

  1. Express as a prod­uct of prime pow­ers: (a) 26952695, (b) 64806480, (c) 93179317.
  2. A num­ber is 24×3×722^4 \times 3 \times 7^2. Write down three dif­fer­ent fac­tors of it that are greater than 100100.
  3. Explain why 6n6^n can never end in the digit 00, for any nat­ural num­ber nn.

Sec­tion B, HCF and LCM

  1. Find the HCF and LCM of 9696 and 404404 by prime fac­tori­sa­tion, then ver­ify HCF×LCM=\mathrm{HCF} \times \mathrm{LCM} = prod­uct.
  2. Given HCF(867,255)=51\mathrm{HCF}(867, 255) = 51, find the LCM with­out fac­toris­ing.
  3. Find the LCM of 88, 99 and 2525.
  4. Three wires of lengths 126126 cm, 180180 cm and 234234 cm are cut into equal pieces of the great­est pos­si­ble length, with noth­ing left over. Find that length and the total num­ber of pieces.
  5. Two lights flash at inter­vals of 4848 sec­onds and 6060 sec­onds respec­tively, and flash together at 8:008{:}00 am. At what time do they next flash together?

Sec­tion C, Irra­tional num­bers

  1. Prove that 11\sqrt{11} is irra­tional.
  2. Prove that 2−532 - 5\sqrt{3} is irra­tional, given that 3\sqrt{3} is irra­tional.
  3. State whether each is ratio­nal or irra­tional, with a rea­son: (a) 16\sqrt{16}, (b) 2×8\sqrt{2} \times \sqrt{8}, (c) 3+3\sqrt{3} + \sqrt{3}.
  4. A stu­dent writes: "The sum of two irra­tional num­bers is always irra­tional." Give a coun­terex­am­ple.

Answers

Show answers

  1. (a) 2695=5×72×112695 = 5 \times 7^2 \times 11. (b) 6480=24×34×56480 = 2^4 \times 3^4 \times 5. (c) 9317=7×1139317 = 7 \times 11^3.

  2. Any three of, for exam­ple, 112=24×7112 = 2^4 \times 7, 147=3×72147 = 3 \times 7^2, 336=24×3×7336 = 2^4 \times 3 \times 7, 784=24×72784 = 2^4 \times 7^2. Each uses only primes avail­able, with expo­nents no larger than those in the num­ber.

  3. End­ing in 00 requires divis­i­bil­ity by 10=2×510 = 2 \times 5, so a fac­tor 55 must appear. But 6n=2n×3n6^n = 2^n \times 3^n con­tains only the primes 22 and 33, and by unique fac­tori­sa­tion no 55 can appear.

  4. 96=25×396 = 2^5 \times 3 and 404=22×101404 = 2^2 \times 101. Com­mon prime 22, smaller expo­nent 222^2, so HCF=4\mathrm{HCF} = 4. All primes, larger expo­nents: LCM=25×3×101=9696\mathrm{LCM} = 2^5 \times 3 \times 101 = 9696. Check: 4×9696=38784=96×4044 \times 9696 = 38784 = 96 \times 404. ✓

  5. LCM=867×25551\displaystyle \mathrm{LCM} = \frac{867 \times 255}{51}. Divide first: 255÷51=5255 \div 51 = 5, so LCM=867×5=4335\mathrm{LCM} = 867 \times 5 = 4335.

  6. 8=238 = 2^3, 9=329 = 3^2, 25=5225 = 5^2 share no prime, so the LCM is the prod­uct: 23×32×52=18002^3 \times 3^2 \times 5^2 = 1800.

  7. 126=2×32×7126 = 2 \times 3^2 \times 7, 180=22×32×5180 = 2^2 \times 3^2 \times 5, 234=2×32×13234 = 2 \times 3^2 \times 13. Com­mon: 2×32=182 \times 3^2 = 18. Great­est length 1818 cm, giv­ing 7+10+13=307 + 10 + 13 = 30 pieces.

  8. 48=24×348 = 2^4 \times 3, 60=22×3×560 = 2^2 \times 3 \times 5, so LCM=24×3×5=240\mathrm{LCM} = 2^4 \times 3 \times 5 = 240 sec­onds =4= 4 min­utes. They next flash together at 8:048{:}04 am.

  9. Assume 11=ab\displaystyle \sqrt{11} = \frac{a}{b} with a,ba, b shar­ing no fac­tor. Then a2=11b2a^2 = 11b^2, so 11∣a211 \mid a^2, and since 1111 is prime, 11∣a11 \mid a. Put a=11ca = 11c: 121c2=11b2121c^2 = 11b^2, so b2=11c2b^2 = 11c^2, giv­ing 11∣b11 \mid b. Both are divis­i­ble by 1111, con­tra­dict­ing the assump­tion.

  10. Sup­pose 2−53=r2 - 5\sqrt{3} = r is ratio­nal. Then 3=2−r5\displaystyle \sqrt{3} = \frac{2 - r}{5}, which is a dif­fer­ence of ratio­nals divided by a ratio­nal, hence ratio­nal. That con­tra­dicts 3\sqrt{3} being irra­tional.

  11. (a) Ratio­nal, 16=4\sqrt{16} = 4. (b) Ratio­nal, 2×8=16=4\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4. (c) Irra­tional, 3+3=23\sqrt{3} + \sqrt{3} = 2\sqrt{3}, a non-zero ratio­nal times an irra­tional.

  12. 5\sqrt{5} and −5-\sqrt{5} are both irra­tional, and their sum is 00, which is ratio­nal. (Equally, 2+72 + \sqrt{7} and 2−72 - \sqrt{7} sum to 44.)