NCERT Solutions: Class 6 Maths Chapter 5 - Prime Time

‘Idli-vada’ is said at common multiples of $3$ and $5$, that is multiples of $15$. The $10$th such number is $10\times15=150$.

From $1$ to $90$, multiples of $3$ are $\lfloor90/3\rfloor=30$, multiples of $5$ are $\lfloor90/5\rfloor=18$, and common multiples of $3$ and $5$ are multiples of $15$, giving $\lfloor90/15\rfloor=6$.

Use the same counting rule up to $900$: $\lfloor900/3\rfloor=300$, $\lfloor900/5\rfloor=180$, and $\lfloor900/15\rfloor=60$.

The overlapping region of the figure represents common multiples. In the game, those common multiples are exactly the numbers where ‘idli-vada’ is said.

If one number is $4$ and nobody says just ‘vada’, every multiple of the other number must also be a multiple of $4$. Among the given choices, $8$ has this property for the game context.

A jump size that reaches both numbers must be a common factor of $15$ and $30$. The factors of $15$ are $1,3,5,15$, and all four also divide $30$.

The shaded entries follow the pattern of multiples of $3$ in the table.

The circled entries follow the pattern of multiples of $4$ in the table.

Numbers that are both shaded and circled are divisible by both $3$ and $4$. In the shown range, these are $36,48,$ and $60$.

Multiples of $40$ near the range are $40\times7=280$, $40\times8=320$, $40\times9=360$, $40\times10=400$, and $40\times11=440$. Those between $310$ and $410$ are $320,360,400$.

For a, multiples of $7$ below $40$ are $7,14,21,28,35$; only $35$ has digit sum $8$. For b, the number must be a multiple of $15$ below $100$; $45$ has digits differing by $1$.

The factors of $6$ are $1,2,3,6$. Their sum is $1+2+3+6=12=2\times6$, so $6$ is perfect.

List the factors of each number and keep only the factors common to all numbers in that part.

Odd multiples of $25$ are not multiples of $50$. Examples are $25\times1=25$, $25\times3=75$, and $25\times5=125$.

The first ‘idli-vada’ is the least common multiple of the two numbers. For $(7,8)$ it is $56$, for $(7,9)$ it is $63$, and for $(8,9)$ it is $72$, all after $50$.

The possible jump sizes are the common factors of $28$ and $70$. Since $28=2\times2\times7$ and $70=2\times5\times7$, the common factors are $1,2,7,14$.

Use the required prime powers for $1,2,3,4,5,6,8,9,10$: $2^3$ from $8$, $3^2$ from $9$, and $5$ from $5$ or $10$. Thus the smallest number is $2^3\times3^2\times5=8\times9\times5=360$.

The least common multiple of $1$ to $10$ uses prime powers $2^3,3^2,5,$ and $7$. Therefore $2^3\times3^2\times5\times7=8\times9\times5\times7=2520$.

The primes from $21$ to $30$ are $23$ and $29$. The remaining eight numbers are composite.

Every even number greater than $2$ is divisible by $2$ and has more than two factors, so it is composite.

The smallest gap occurs between $2$ and $3$: $3-2=1$. The largest gap among successive primes below $100$ occurs between $89$ and $97$: $97-89=8$.

Count the primes in each decade of the table. The counts are not equal across rows.

$23$ and $37$ have no factors other than $1$ and themselves. $51=3\times17$ and $26=2\times13$ are composite.

$2+3=5$, $3+7=10$, and $2+13=15$, all multiples of $5$.

Each pair uses the same two digits in reverse order, and both numbers in each pair are prime.

Each of these seven numbers is composite: $90,92,94,96$ are even, $91=7\times13$, $93=3\times31$, and $95=5\times19$.

Check consecutive prime pairs below $100$ and select the pairs whose difference is $2$.

a. A number ending in $4$ is even and greater than $2$, so it is composite. b. A product of primes has those primes as factors, so it is composite. c. A prime has exactly two factors, $1$ and itself. d. $2$ is even and prime. e. After any odd prime, the next number is even and greater than $2$, hence composite.

Any number ending in $2$ or $4$ is even, and any number ending in $5$ is divisible by $5$. Since each arrangement ends in $2$, $4$, or $5$, no arrangement is prime.

For each prime $p$, compute $2p+1$: $2\times5+1=11$, $2\times11+1=23$, $2\times23+1=47$, $2\times29+1=59$, and $2\times41+1=83$, all prime.

A pair is safe when the two numbers are co-prime, so no jump size greater than $1$ reaches both. $4$ and $15$ share no factor greater than $1$, and $18$ and $29$ share no factor greater than $1$.

Each listed co-prime pair has no common factor greater than $1$. The pairs $30$ and $415$ share $5$, and $81$ and $18$ share $9$, so they are not co-prime.

For co-prime pairs, the first common multiple is the product of the two numbers. For pairs with a common factor greater than $1$, the first common multiple is less than the product.

Divide each number successively by prime factors until only primes remain. The listed products multiply back to the original numbers.

The number is $2\times3\times3\times11=198$.

$1955\div5=391$, and $391=17\times23$. Thus $1955=5\times17\times23$, with all three primes less than $30$.

Factor each component first: $56=2^3\times7$, $25=5^2$; $108=2^2\times3^3$, $75=3\times5^2$; $1000=2^3\times5^3$, $81=3^4$.

For the smallest such numbers, use the smallest different primes. Three different primes give $2\times3\times5=30$; four different primes give $2\times3\times5\times7=210$.

$30=2\times3\times5$ and $45=3^2\times5$ share factors. $57=3\times19$ and $85=5\times17$ share none. $121=11^2$ and $1331=11^3$ share $11$. $343=7^3$ and $216=2^3\times3^3$ share none.

$225=3^2\times5^2$ does not contain $27=3^3$. $96=2^5\times3$ contains $24=2^3\times3$. $343=7^3$ does not contain factor $17$. $999=3^3\times37$ does not contain $99=3^2\times11$.

Both numbers share factors $3$ and $7$, so they are not co-prime. The first has factor $2$ which the second lacks, and the second has factor $11$ which the first lacks, so neither prime factorisation is contained in the other.

Two distinct primes have no common factor other than $1$, so they are co-prime. For example, $2$ and $3$, or $3$ and $11$.

Only the last two digits need to be checked for divisibility by $4$. Since $36$ is divisible by $4$, $8536$ is divisible by $4$.

Any whole number can be split as a multiple of $100$ plus its last two digits. Since $100$ is divisible by $4$, divisibility by $4$ depends only on the number formed by the last two digits.

The numbers in each range whose last three digits form a multiple of $8$ are listed. This suggests that the last three digits decide divisibility by $8$.

$8552\div8=1069$, so $8552$ is a multiple of $8$.

Any whole number can be split as a multiple of $1000$ plus its last three digits. Since $1000$ is divisible by $8$, divisibility by $8$ depends only on the number formed by the last three digits.

For part b, count multiples of $4$ from $2024$ through $2099$: $2024,2028,\ldots,2096$. This arithmetic sequence has $((2096-2024)/4)+1=19$ terms. The year $2100$ is outside the range.

For even numbers, $2+6=8$ is a multiple of $4$, but $2+4=6$ is not. For odd numbers, $1+3=4$ is a multiple of $4$, but $1+5=6$ is not.

For division by $10$, the remainder is the last digit. For division by $5$, reduce the last digit modulo $5$. For division by $2$, even numbers have remainder $0$ and odd numbers have remainder $1$.

If a number is divisible by $8$, it is divisible by $2$ and $4$. If it is also divisible by $5$, then being even makes it divisible by $10$. Since $14560$ is divisible by both $5$ and $8$, it is divisible by all five numbers.

A number divisible by all of $2,4,5,8,10$ must be divisible by $8$ and end in $0$. The numbers $5600$, $6000$, and $77622160$ satisfy this; $572$ and $2352$ do not end in $0$.

$16\times625=10000$, and neither $16$ nor $625$ has $0$ as its units digit.