Home work solution for Class 1 | Divisibility Rules (Part 1)| Calculation Technique | Number System | By Nitesh Sir
Let’s check divisibility of 2, 3, 4, 5, 6, 8, 9, 10 and 11 for each given number step-by-step.
ЁЯФв 1) Number = 7,843,218
| Divisor | Test | Result |
|---|---|---|
| 2 | Last digit = 8 (even) | ✅ Divisible |
| 3 | Sum = 7+8+4+3+2+1+8 = 33 | ✅ (33 divisible by 3) |
| 4 | Last 2 digits = 18 | ❌ |
| 5 | Last digit ≠ 0 or 5 | ❌ |
| 6 | Divisible by 2 & 3 | ✅ |
| 8 | Last 3 digits = 218 | ❌ |
| 9 | Sum = 33 | ❌ |
| 10 | Last digit ≠ 0 | ❌ |
| 11 | (7+4+2+8) − (8+3+1) = 21 − 12 = 9 | ❌ |
✔ Divisible by: 2, 3, 6
ЁЯФв 2) Number = 7,892,846
| Divisor | Test | Result |
|---|---|---|
| 2 | Last digit = 6 | ✅ |
| 3 | Sum = 7+8+9+2+8+4+6 = 44 | ❌ |
| 4 | Last 2 digits = 46 | ❌ |
| 5 | Last digit ≠ 0 or 5 | ❌ |
| 6 | Needs 2 & 3 | ❌ |
| 8 | Last 3 digits = 846 | ❌ |
| 9 | Sum = 44 | ❌ |
| 10 | Last digit ≠ 0 | ❌ |
| 11 | (7+9+8+6) − (8+2+4) = 30 − 14 = 16 | ❌ |
✔ Divisible by: 2 only
ЁЯФв 3) Number = 93,475,116
| Divisor | Test | Result |
|---|---|---|
| 2 | Last digit = 6 | ✅ |
| 3 | Sum = 9+3+4+7+5+1+1+6 = 36 | ✅ |
| 4 | Last 2 digits = 16 | ✅ |
| 5 | Last digit ≠ 0 or 5 | ❌ |
| 6 | Divisible by 2 & 3 | ✅ |
| 8 | Last 3 digits = 116 | ❌ |
| 9 | Sum = 36 | ✅ |
| 10 | Last digit ≠ 0 | ❌ |
| 11 | (9+4+5+1) − (3+7+1+6) = 19 − 17 = 2 | ❌ |
✔ Divisible by: 2, 3, 4, 6, 9
ЁЯУМ Final Summary
| Number | Divisible by |
|---|---|
| 7,843,218 | 2, 3, 6 |
| 7,892,846 | 2 |
| 93,475,116 | 2, 3, 4, 6, 9 |