NCERT Solutions · Class 9 Mathematics The World of Numbers

Unitary method — scale a fixed rate.The exchange rate is fixed: \(\displaystyle 15\) ingots for every \(\displaystyle 2\) bags. So the answer only depends on how many lots of \(\displaystyle 2\) bags he brings.\[12 \text{ bags} \div 2 \text{ bags per lot} = 6 \text{ lots} \]Each lot is worth \(\displaystyle 15\) ingots, so\[6 \times 15 = 90 \text{ ingots.} \]Check by a second route (rate per single bag). One bag fetches \(\displaystyle \frac{15}{2} = 7.5\) ingots, so \(\displaystyle 12\) bags fetch\[12 \times \frac{15}{2} = \frac{180}{2} = 90. \]The two routes agree. Notice that the rate per bag is not a whole number, yet the total is — because \(\displaystyle 12\) is a multiple of \(\displaystyle 2\), the halves pair up.He will leave with $\displaystyle 90$ copper ingots.

Test each number for factors.Check whether each number has any divisor other than \(\displaystyle 1\) and itself. It is enough to test the prime divisors up to the square root of the number: a factor bigger than the square root must be paired with one smaller, and any composite factor is itself built from smaller primes. All four numbers here are below \(\displaystyle 25 = 5^2\), so only \(\displaystyle 2\) and \(\displaystyle 3\) need to be tried.

Number	Divisible by \(\displaystyle 2\)?	Divisible by \(\displaystyle 3\)?	Verdict
\(\displaystyle 11\)	no (odd)	no (\(\displaystyle 1+1 = 2\))	prime
\(\displaystyle 13\)	no (odd)	no (\(\displaystyle 1+3 = 4\))	prime
\(\displaystyle 17\)	no (odd)	no (\(\displaystyle 1+7 = 8\))	prime
\(\displaystyle 19\)	no (odd)	no (\(\displaystyle 1+9 = 10\))	prime

(The digit-sum test: a number is a multiple of \(\displaystyle 3\) exactly when its digits add to a multiple of \(\displaystyle 3\).)So every one of them is a prime number. In fact these four are all the primes between \(\displaystyle 10\) and \(\displaystyle 20\) — that is what makes this column of the Ishango bone so striking. (They also form two "twin" pairs, \(\displaystyle 11, 13\) and \(\displaystyle 17, 19\), each pair two apart.)Continuing the pattern. Take the numbers after \(\displaystyle 19\) in turn: All four are prime numbers. The next three primes are $\displaystyle 23$, $\displaystyle 29$ and 31.

Closure is broken by a single counterexample."Closed under an operation" means: take any two members of the set, apply the operation, and the result must land back inside the same set. Because the claim is about every pair, one failing pair is enough to disprove it.The natural numbers are \(\displaystyle 1, 2, 3, 4, \ldots\)Example 1. \(\displaystyle 3\) and \(\displaystyle 8\) are both natural numbers, but\[3 - 8 = -5, \]and \(\displaystyle -5\) is not a natural number — you cannot reach it by counting.Example 2. \(\displaystyle 7\) and \(\displaystyle 7\) are both natural numbers, but\[7 - 7 = 0, \]and \(\displaystyle 0\) is not a natural number either (counting starts at \(\displaystyle 1\)).Why it fails in general. For natural numbers \(\displaystyle a\) and \(\displaystyle b\), the difference \(\displaystyle a - b\) is again natural only when \(\displaystyle a > b\). Since subtraction is allowed to take the numbers in either order, the set cannot be closed.This is exactly the gap that pushed mathematicians to invent \(\displaystyle 0\) and the negative numbers; enlarging the set to the integers \(\displaystyle \ldots, -2, -1, 0, 1, 2, \ldots\) makes subtraction always possible, because every difference then lands back inside the set.(Your own counterexamples may differ from the ones above — any pair with \(\displaystyle a \le b\) works equally well. The conclusion, though, is the same for everyone.)No — the natural numbers are not closed under subtraction.

Count the joints, not the fingers.The thumb is doing the pointing, so it is the counter, not something counted. That leaves four fingers, and each of those has \(\displaystyle 3\) joints (the creases where the finger bends):\[4 \times 3 = 12. \]You can count up to $\displaystyle 12$ on one hand.How this relates to base-$\displaystyle 12$ (duodecimal) counting. A base is just the size of the "bundle" you use before you start again. Counting on the finger joints of one hand naturally produces bundles of \(\displaystyle 12\), not \(\displaystyle 10\) — which is why so many old trading systems counted in dozens: $\displaystyle 12$ items to a dozen, $\displaystyle 12$ months in a year, $\displaystyle 12$ inches in a foot, $\displaystyle 12$ hours on a clock face.Twelve also happens to be a very convenient base for a merchant, because it divides evenly in many ways: \(\displaystyle 12 = 2 \times 6 = 3 \times 4\), so a dozen can be split into halves, thirds, quarters and sixths without breaking anything. Ten only splits into halves and fifths.And how base $\displaystyle 60$ appears. Use the other hand to keep track of how many complete dozens you have counted. Five fingers on that hand means five dozens:\[5 \times 12 = 60. \]That is one reason the Babylonian base-$\displaystyle 60$ system took hold, and its traces are still with us: \(\displaystyle 60\) seconds in a minute, \(\displaystyle 60\) minutes in an hour, and \(\displaystyle 360 = 6 \times 60\) degrees in a full turn.One hand gives $\displaystyle 12$ counts — the origin of the dozen and of base-$\displaystyle 12$; with the second hand tallying dozens you reach $\displaystyle 60$, the origin of base-60.