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In order to solve difficulties based on Number Systems, dividing, or more precisely, divisibility principles, can be quite helpful at times. The conventional rules for 2n and 5n are those with which almost all of us feel most at ease. All that is required to determine these is to glance at the last 'n' digits of the number. A number is divisible by 2n or 5n if its last 'n' digits are also divisible by 2n or 5n, and vice versa.
Relationship between Remainder and Decimal
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Let's say we divide 42 by 5. The outcome has an 8-quotient and a 2-remainder. But 42/5 Equals 8.4. The answer has an integer part and a decimal part, as you can see. The decimal portion is 0.4 and is determined by 2/5, with the integer portion being 8 (which equals the quotient).
We should be aware that any divisor N will produce exactly N possible remainders because we have previously learned that the set of remainders for any divisor N obeys the inequality 0 $ R N. (For instance, there are three potential remainders (0, 1, and 2) if the divisor is 3.
Theorems of Divisibility
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- If a can be divided by b, then ac can be divided by b as well.
- A is divisible by c if and only if b and c are both divisible by c.
- If a and b are both natural numbers such that a and b are divisible by each other, then a = b.
- If (m + n) and (m - n) are both divisible by d, then n and m must both be divisible by d. There is a significant implication here. Let's say that 7 is the sum of the divisors of 28 and 742. Therefore, both (742 + 28) and (742 - 28) can be divided by 7. (And + 28 - 742 is correct, too.)
- A is divisible by bd if an is divisible by b, and c is divisible by d if c is divisible by d.
- The highest power of a prime number, [n/p] + [n/p2] + [n/p3] +..., which divides n! precisely. [x] signifies the largest integer less than or equal to x.
Short Cuts
- Whether a number can divisible by 2 or 5 depends on whether the final digit is divisible by 2 or 5.
- All such integers that have digit sums that really are divisible by 3 (or 9) are all divisible by 3 (or 9).
- An integer is said to be divisible by 4 if its final two digits are also divisible by 4.
- Whether a number is divisible by 6 depends on whether it is also simultaneously divisible by 2 and 3.
- An integer is said to be divisible by 8 if its final three digits are also divisible by 8.
- A number is considered to be divisible by 11 if the total of the odd and even digits divided by the sum of the odd and even digits is zero or divisible by 11.
- All numbers that are divisible by 3 and 4 are also divisible by 12.
- The number n is divisible by 7, 11, or 13 if and only if the residue of its division by 1000, after subtracting the number of its thousands, is also divisible by 7, 11, or 13.
Sample Questions
Ques. When you divide - 24.8 by 6, what is left over?
Ans. – 24.8 = 6 * (– 4) + (– 0.8)
Remainder in the negative, not correct.
–24.8 = 6 * (–5) + 5.2
Positive value for the remainder, which is true.
Ques. When a long piece of wood is divided into equal parts, each measuring 242 cm, a small piece, measuring 98 cm, remains. The length of the wood that would remain if this Part were divided into equal pieces, each measuring 22 cm, would be
Ans. The length of wood needed will equal the remaining after 98 is divided by 22 because 242 is divisible by 22:
As a result, 10 [98/22; remaining 10]
Ques. Determine how many numbers between 1 and 100 are not divisible by 2.
Ans. The first non-divisible by two number from 1 to 100 is 1, and the last non-divisible by two number from 1 to 100 is 99.
From 1 to 99, no alternate number (i.e., at the gap of 2) will be divisible by 2. (1, 2, 3, - - - -, 95, 97, 99)
Therefore, the needed number of no’s = ((last no. - first no.)/ gap (or step)) +1
= ((99-1)/2)) + 1 = 50
Ques. Find all 34x5y five-digit numbers that have divisors of 36.
Ans. 36 is the result of the coprime numbers 4 and 9. Therefore, 34x5y will be divisible by 36 if it is divisible by 4 and 9. Consequently, we get that the value of y can be either 2 or 6. Additionally, the value becomes 34x52 if y equals 2. The sum of 3 + 4 + x + 5 + 2 should divide by 9 for this to be divisible by 9. Due to this, x may be 4.
Therefore, 36 can be divided by the number 34452.
When the sum of the digits is divisible by 9, the number 34x56 will also be divisible by 36 for y = 6. When x is either 0 or 9, this will occur. 34056 and 34956 will therefore be divisible by 36.
Ques. Multiplying 5 x 10 x 15 x 20 x 25 x 30 x 35 x 40 x 45 x 50, get the number of zeros.
Ans. The quantity of zeros is influenced by the quantity of fives and twos. Close examination reveals that the constraint in this case is the number of twos. The phrase might be expressed as
Number of 5s – 12, Number of 2s – 8.
Hence: 8 zeroes.
Ques. A positive integer K must be chosen such that k + 4 is divisible by 7. Then, the smallest positive number n that is greater than 2 and divisible by 7 equals k + 2n.
Ans. A number of the form 7n + 3 will be k. Therefore, k + 2n becomes 7n + 3 + 18 = 7n + 21 if n is taken to be 9 instead. This sum can be divided by 7. We are unable to arrive at this conclusion with the numbers 3, 5, or 7.
Ques. What precise power of 8 divides 25?
Ans. If the integer 8 were a prime, the solution would be [25/8] = 3. Use the subsequent procedure though, as 8 is not a prime number.
2 2 2 and 2 are the prime factors of 8. We require three twos in order to divide by 8. Therefore, whenever we find 3 pairs of 2, we add 1 to the power of 8 to divide 25! We do the following to count how we end up with 3 twos. Every even number will result in at least one "two" [25/2] = 12
Additionally, all 25! numbers that are divisible by 22 will result in a bonus two [25/22] = 6
In addition, any integer in 25! that is divisible by 23 results in a third two. Hence [25! /23] = 3
And every integer in 25 that can divide by 24 will result in a fourth two. Hence [25! /24] = 1
Thus, there are 22 twos in all in the number 25. We require three twos to divide a number by 8. As a result, 25! will be divided by 8 seven times because it contains 22 twos.
Ques. What fraction of a power of 15 is 87?
Ans. 15 Equals 5 * 3. As a result, we shall count one whenever we can make a pair of one 5 and one 3. 87! has [87/5] + [87/52], which adds up to 17 + 3 = 20 fives.
Additionally, the number 87! contains [87/3] + [87/32] + [87/33] + [87/34] = 29 +... (more than 20 threes).
Therefore, 15 will divide 87! twenty times because the number of 5s, not the number of 3s, is what limits the power.
In truth, it is not difficult to see that when all of the factors are prime, we only need to find the greatest prime number to serve as a limit on the denominator's power. Therefore, in this instance, we only needed to check for the number of 5s.
Ques. Determine how many numbers between 300 and 400 (both inclusive) are not divisible by 2, 3, 4, or 5.
Ans. 101 numerals altogether
Step 1: Not divisible by two = All even integers are rejected: 51
50 are still standing.
Step 2: Of which the first number, 300, and the last number, 399, are divisible by 3. Only odd numbers between 300 and 400 that are divisible by three remain because even numbers have previously been eliminated. This provides us with: The first two numbers, 303 and 399, have a common difference. 6
Therefore, eliminate: [(399 - 303)/6] + 1 = 17; 50 - 17; 33 numbers remain.
Since only even numbers will be left after removing all terms that are divisible by 4, we do not need to remove any further terms (which have been obliterated).
Step 3: From 33 numbers, remove All odd numbers that may be divided by 5 but not by 3 were left. The first odd integer divisible by five between 300 and 400 is 305, and the last is 395. Because both ends are tallied, we get 10 such numbers, as in [(395 - 305)/10 + 1 = 10].
To get to 33 digits, some of these 10 numbers have previously been dropped. Operation left: decrease all of the remaining 10 numbers, including 305, 315, and 395, that are also divisible by 3. A quick glance reveals that the numbers all begin with 315 and share the difference of 30.
Hence (375 - 315)/30 + 1 = [(Last number - First number)/Difference + 1]
The initial 100 had already been reduced by these three numbers. Therefore, for integers that can be divided by 5, we simply need to exclude the odd numbers that can also be divided by 5 but not by 3. Between 300 and 400, there are seven such numbers. the remaining numbers are: 33 - 7 = 26.
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