Number Play Class 8 Solutions

If one-term sums are allowed, every natural number is its own sum. Using at least two positive consecutive integers, exactly the positive integers that are not powers of 2 can be represented. Every odd number greater than 1 is (n-1)/2+(n+1)/2. Some even numbers also work, such as 10=1+2+3+4, but powers of 2 do not.

With negative integers, 0=(-1)+0+1.

The eight values are: 3+4+5+6=18, 3+4+5-6=6, 3+4-5+6=8, 3+4-5-6=-4, 3-4+5+6=10, 3-4+5-6=-2, 3-4-5+6=0, and 3-4-5-6=-12. All are even.

Every result is even. Writing the numbers as n,n+1,n+2,n+3, any choice of signs has the same parity as their all-plus sum 4n+6, which is even.

All eight results have the same parity. Each sign change changes the value by twice an integer, so it cannot change parity. Their common parity is the parity of a+b+c+d.

Changing -c to +c increases the value by 2c; changing -d to +d increases it by 2d. The change is always even, so parity stays unchanged.

Switching the sign of a number replaces its tokens by the same number of opposite tokens. The net value changes by two times that number, an even amount. Therefore no sign switch changes parity.

No. For any fixed list of integers, changing any selection of plus and minus signs changes the value by a sum of terms of the form 2k. Hence every sign arrangement has the same parity, regardless of how many numbers are used.

The even expressions are 43+37, 672-348, 4\times347\times3, and 543-479. The other four are odd: 708-477, 809+214, 119\times303, and 513^3.

Always even: 2a+2b, 4m+2n, 2u-4v, 13k-5k=8k, and 4k\times3j=12kj. The expressions 3g+5h, 6m-3n, x^2+2, and b^2+1 are not always even.

An expression is guaranteed even when 2 can be factored from every term. Examples include 6r+10s=2(3r+5s) and 4x^2-2y=2(2x^2-y). For a non-example such as 3g+5h, g=h=1 gives 8 but g=1,h=2 gives 13.

Write each even number as either 4p or 4p+2. The sum is divisible by 4 when both have the same type: 4p+4q, or (4p+2)+(4q+2). If one has each type, the sum leaves remainder 2.

4p+(4q+2)=4(p+q)+2. It is even but not divisible by 4; for example, 12+6=18.

Always true. If the numbers are 8a and 8b, their difference is 8(a-b), a multiple of 8.

Sometimes true. For example, 72=48+24 uses two multiples of 8, while 72=50+22 does not.

Always true. If A=7j, then any multiple mA=7(jm).

Always true. Writing it as 12m shows that 1, 2, 3, 4, 6, and 12 all divide it.

Sometimes true. For example, 42 is divisible by 14 but not by 28. In general, 7k is divisible by 7m exactly when m divides k.

Always true because \operatorname{lcm}(9,4)=36.

Sometimes true. Such a number must be divisible by \operatorname{lcm}(6,4)=12, but not necessarily 24: 12 is a counterexample, while 24 is an example.

Never. Odd plus even is odd, whereas every multiple of 6 is even.

Both 5k+3 for k\ge0 and 5k-2 for k\ge1 generate 3,8,13,18,\ldots. Thus choices (iv) and (v) work.

Equivalent forms include 5(k+1)-2, 5k+8, and 10k+3 together with 10k+8 when the two latter families are combined. The simplest complete form is 5k+3.

The numbers are 7, 8, 9, and 10.

They are p-1,p-2,p-3,p-4.

(i) Sometimes: 2+4=6, but 2+6=8. (ii) Sometimes: 30 is divisible by neither 18 nor 9, but 27 is divisible by 9 and not 18. (iii) Sometimes: 8+10=18, while 9+11=20. (iv) Always: 6x+9y=3(2x+3y). (v) Sometimes: 18+9=27, but 12+9=21.

Since x-2 must be divisible by both 3 and 4, x=12n+2. Examples are 14, 26, 38, and 50.

The count is odd and leaves remainder 1 modulo 3 and 5, so it is 30k+1: 31, 61, or 91 below 100. Only 91 is divisible by 7. Therefore there are 91 pebbles.

Yes. (6a+2)+(6b+2)+(6c+2)=6(a+b+c+1), which is always divisible by 6.

Since 4779=7p+5 and 661=7q+3, the sum is 7(p+q+1)+1, remainder 1. The difference is 7(p-q)+2, remainder 2.

Each remainder is one less than its divisor, so the number plus 1 is a common multiple of 3, 4, and 5. The smallest is \operatorname{lcm}(3,4,5)-1=60-1=59.

All place values above units are multiples of 10, so divisibility by 2 or 5 depends only on the units digit. All place values from 100 upward are multiples of 4, so only the last two digits matter. All place values from 1000 upward are multiples of 8, so only the last three digits matter.

All five are divisible by 9; their digit sums are 27, 18, 9, 9, and 18.

Yes. Every non-zero expanded-place-value term contains a factor of 9, and equivalently the digit sum is a multiple of 9.

Because every power of 10 leaves remainder 1 modulo 9, the remainder equals the digit-sum remainder: 4+2+7=13, and 13\equiv4\pmod9.

All four are correct. A number and its digit sum have the same remainder modulo 9, so each implication and its contrapositive holds.

Only 405 is divisible by 9. The digit sums are 6, 9, 32, 28, and 30 respectively.

It is 288. A digit sum made only from non-zero even digits cannot be 9, so the smallest feasible sum is 18; the smallest arrangement using allowed digits is 288.

It is 6003, which is 3 away. The preceding multiple, 5994, is 6 away.

There are 11: from 4302 through 4392.

Every power of 10 leaves remainder 1 modulo 3. Therefore a number has the same remainder modulo 3 as the sum of its digits.

The alternating sum is 2-6+4=0, a multiple of 11, so 462 is divisible by 11: 462=11\times42. In general, a number is divisible by 11 when its alternating digit sum is a multiple of 11.

It means the original number leaves remainder 0 and is divisible by 11.

The first four are not divisible; their remainders are 4, 5, 8, and 7. The numbers 90904 and 857076 are divisible by 11.

Divisors that work: 128: 2,4,8; 990: 2,3,5,6,9,10,11; 1586: 2; 275: 5,11; 6686: 2; 639210: 2,3,5,6,10,11; 429714: 2,3,6,9; 2856: 2,3,4,6,8; 3060: 2,3,4,5,6,9,10; 406839: 3.

Yes, because 2 and 3 are coprime and \operatorname{lcm}(2,3)=6. Of 38, 225, 186, and 64, only 186 is divisible by both 2 and 3, hence by 6.

Since 3 and 8 are coprime, their LCM is 24. But \operatorname{lcm}(4,6)=12, so passing both tests guarantees only a multiple of 12; 12 itself is the counterexample.

The digital root has the same remainder modulo 9 as the number, except that remainder 0 is written as digital root 9 for a positive number.

Root 5: 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698. Root 7: 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691. Root 3: 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696.

They cycle through 1,2,3,4,5,6,7,8,9 and then repeat. Any block of 12 consecutive integers therefore contains a complete nine-root cycle followed by the first three roots of the next cycle, shifted according to its starting number.

Multiples of 3 cycle 3,6,9. Multiples of 4 cycle 4,8,3,7,2,6,1,5,9. Multiples of 6 cycle 6,3,9.

They cycle 7,4,1 because the multiples of 6 cycle through roots 6,3,9 and adding 1 shifts these to 7,4,1.

The number is 11,11,11,111: it has nine digits, all digits are odd, their sum is 9, and its digital root is 9. Its name is eleven crore eleven lakh eleven thousand one hundred eleven.

Since 10 has digital root 1, the new digital root is 6.

Each addition changes the digital root by 2 modulo 9. For a starting root of 1, the cycle is 1,3,5,7,9,2,4,6,8 and then repeats.

9a+36b+13=9(a+4b+1)+4, so its digital root is 4.

Parity has no fixed relation to digital root. Modulo 3, roots 1,4,7 give remainder 1; 2,5,8 give remainder 2; 3,6,9 give remainder 0. Modulo 9, roots 1 through 8 equal the remainder, while root 9 means remainder 0.

(i) A=7,B=9. (ii) A=2,B=5. (iii) N=1,O=3,P=9. (iv) Q=8,R=5,P=2.

PQ=12 and RS=96, so P=1,Q=2,R=9,S=6. Any larger eligible two-digit input makes a three-digit product.

The valid option is 24\times4=96, so G=2,H=4,K=6.

The hundreds digit forces B=1. The product must remain three-digit, so Y<7, and its occurrence as the units digit of a product by 6 makes it even. Thus the initial candidates are 0, 2, 4, and 6, subject to the remaining column and distinct-letter constraints.

(i) U=5,T=0,P=1. (ii) A=1,B=9,C=5. (iii) L=1,N=5,P=0. (iv) X=2,Y=3,Z=9. (v) P=2,Q=1,R=4. (vi) J=7,K=4.

The digit sum is 9+z, so z=0 or z=9. Both make the sum a multiple of 9.

The sum is (12n+8)+(12m-4)=12(n+m)+4. It is a multiple of 8 only for some values of n+m, not always. For example, 8+8=16 works, but 20+8=28 does not.

Write them as 3m and 3n. Their sum is 3(m+n), which is divisible by 6 exactly when m+n is even, meaning the two multipliers have the same parity.

Yes. Every rearrangement has the same digit sum, so it remains divisible by 9, provided leading-zero conventions are handled when interpreting the result as a numeral.

Divisibility by 2 makes b even, and divisibility by 9 requires 17+a+b to be a multiple of 9. The pairs are (1,0),(8,2),(6,4),(4,6),(2,8).

The last two digits force q\in\{0,2,4,6,8\}. The divisibility-by-11 condition gives 18-(p+q)=0 or 11. The valid pairs are (7,0),(5,2),(3,4),(1,6).

One triple is 2,3,4. The conditions repeat modulo \operatorname{lcm}(2,3,4)=12, so all triples are 12k+2,12k+3,12k+4.

One valid list is 45,036; 45,072; 45,108; 45,144; and 45,180. Consecutive multiples differ by 36.

They are 5p-4,5p-2,5p+2,5p+4. Since the middle number is even, p must be even.

One example is 200025. It ends in 5 and has digit sum 9, so it is divisible by 15. Its reversal, 520002, is even and has the same digit sum 9, so it is divisible by 6.

The claim is false. If n=11k, then 2n=22k=11(2k), so every doubled multiple of 11 remains a multiple of 11.

(i) Always: (6m)(3n)=18mn, a multiple of 9. (ii) Always: (2n-2)+2n+(2n+2)=6n. (iii) Always: rearranging digits preserves digit sum and the final digit remains even, so divisibility by 6 remains. (iv) Never: the expression simplifies to 12b-28, which leaves remainder 8 modulo 12.

Classify each number by remainder 0, 1, or 2 modulo 3. The sum is divisible by 3 when the remainders are all equal, or when they are one each of 0, 1, and 2.

Two consecutive integers have product divisible by 2; three have product divisible by 6; four have product divisible by 24; and five have product divisible by 120. Each block supplies all prime factors in the corresponding factorial.

For the first, E=3,F=7,G=1, giving 37\times3=111. For the second, W=5,O=7,M=2,E=8, giving 575\times5=2875.

Diagram (iv). Every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4, so the sets must be nested in that order.