Euclid’s Division Lemma

Step 1: (37 \times 6=222) and (37 \times 7=259). Step 2: (259) is greater than (250), so the quotient is (6). Step 3: Checking the next multiple helps choose the correct quotient.

Step 1: From the previous calculation, (q=6). Step 2: (r=250-37 \times 6=250-222=28). Step 3: After getting the answer, check that (28<37).

Step 1: A number exactly divisible by (10) has the form (10q). Step 2: Multiples of (10) end in (0). Step 3: Euclidean forms can also be used in digit-based questions.

Step 1: If division by (2) leaves remainder (1), the number has the form (2q+1). Step 2: Such a number is odd. Step 3: Not every odd number is prime or a perfect square, so avoid quick assumptions.

Step 1: (n=2q) means (n) is exactly divisible by (2). Step 2: Therefore, (n) is an even number. Step 3: The form (2q) is very useful for identifying even numbers.

Step 1: The remainder must be smaller than (31), but (32) is larger. Step 2: (31q+32=31(q+1)+1), so the remainder is (1). Step 3: If the given form has a large remainder, rewrite it correctly.

Step 1: The remainder is always smaller than the divisor. Step 2: On division by (12), the greatest possible remainder is (12-1=11). Step 3: When asked for the greatest remainder, think of (b-1).

Step 1: The remainder can start from (0). Step 2: If the number is exactly divisible by (12), the remainder is (0). Step 3: The smallest possible remainder is always (0).

Step 1: (25 \times 17=425). Step 2: (432-425=7), so the remainder is (7). Step 3: In questions involving (25), nearby multiples are easy to find, so use them.

Step 1: Check multiples of (7). Step 2: (7 \times 9=63) and (7 \times 10=70), which is greater than (64). So (q=9). Step 3: Take the quotient for which the product does not exceed the number.

Step 1: Taking (q=9), (7 \times 9=63). Step 2: (64-63=1), so (r=1). Step 3: After finding the quotient, subtract to get the remainder.

Step 1: In the correct form, (0 \le r < 9). Step 2: In (58=9 \times 6+4), the remainder is (4), and (4<9). Step 3: Along with equality, the range of the remainder must also be correct.

Step 1: (20) cannot be the remainder because it is greater than (18). Step 2: (20=18+2), so (18q+20=18(q+1)+2). Step 3: The correct remainder always lies from (0) to one less than the divisor.

Step 1: (100 \times 9=900). Step 2: (999-900=99), so the remainder is (99). Step 3: In division by (100), the last two digits often give the remainder, but still check (r<100).

Step 1: When (r=0), the form becomes (a=bq). Step 2: This means (a) is a multiple of (b). Step 3: Zero remainder indicates exact divisibility.

Step 1: Dividing (37) by (37), it divides exactly once. Step 2: So (37=37 \times 1+0), hence (q=1) and (r=0). Step 3: When dividend and divisor are equal, the remainder is (0).

Step 1: When the dividend is smaller than the divisor, the quotient is (0). Step 2: (29=50 \times 0+29), and (29<50), so it is correct. Step 3: When a smaller number is divided by a larger number, the remainder can be the smaller number itself.

Step 1: On division by (7), the possible remainders are (0,1,2,3,4,5,6). Step 2: There are (7) possible remainders in total. Step 3: For a divisor (b), the number of possible remainders is (b).

Step 1: Write (n=6q+5). Step 2: (n+1=6q+6=6(q+1)+0), so the remainder is (0). Step 3: If the remainder is (b-1) and (1) is added, the new remainder becomes (0).

Step 1: Write (n=9q+2). Step 2: (n+5=9q+7), so the remainder on division by (9) is (7). Step 3: Add the added number to the remainder; if the sum is smaller than the divisor, it becomes the new remainder.