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Alik [6]
2 years ago
11

6. Fill in the smallest digit to make the number divisible by:

Mathematics
1 answer:
Degger [83]2 years ago
7 0

Answer:

(i) 7164<u>0</u>, 32197<u>0</u>

(ii) 1<u>1</u>43, 47<u>2</u>05, <u>2</u>316

(iii) <u>1</u>428, 9<u>2</u>52, 721<u>2</u>

(iv) 2462<u>0</u>, 91<u>00</u>, 670<u>0</u>

(v) 1232<u>0</u>, 59<u>0</u>16, 4642<u>4</u>

Step-by-step explanation:

(i) In order for any number to be divisible by 5, its last digit must be <u>either 0 or 5</u>.

In this case, we should add a last digit, so we opt for <u>0</u> without any doubt since it gives a smaller number than <u>5</u> as the last digit: 71640 is smaller than 71645, the same way as 321970 is smaller than our second option, 321975.

(ii) If the sum of a number's digits gives a number divisible by 3, then the main number is also divisible by 3.

This means that we have to do some addition:

1_43: 1 + 4 + 3 = 8. Eight is not divisible by 3. But <u>9</u> is. So the smallest digit we can add for it to be divisible by three is <u>1</u>: <em>1143.</em>

47_05: 4 + 7 + 0 + 5 = 16. Since 16 is not, the next number that is divisible by 3 is 18, and therefore, the smallest digit we need here is <u>2</u>: <em>47205.</em>

_316: 3 + 1 + 6 = 10. Twelve is the next number divisible by 3, and that means <u>2</u> is the digit we choose: <em>2316.</em>

(iii) Only numbers that are divisible by both <u>2 and 3</u> are also divisible by 6.

_428: In order for a number to be divisible by 2, its last digit must be 0 or divisible by 2. That is already the case here.

4 + 2 + 8 = 14 which is not divisible by 3. But 15 is. So the smallest digit we need is <u>1</u>: <em>1428.</em>

9_52: The number will be divisible by 2. And 9 + 5 + 2 = 16 which is not divisible by 3 but 18 is. This means the smallest digit we need is <u>2</u>: <em>9252.</em>

721_: Since the last digit can be any of these <u>0, 2, 4, 6, 8</u> for the number to be divisible by two, we will first see what digit we need for it to be divisible by 3.

7 + 2 + 1 = 10 which is not divisible by 3 but 12 is. We will then choose to add <u>2</u>: <em>7212.</em>

(iv) A number is divisible by 4 only if the number that is formed by its two last digits is also divisible by 4..

2462_: Since we need to add a last digit, this should be easy. The smallest two-digit-number that starts with 2 and is divisible by four is <u>20</u>: <em>24620. </em>

91__: Here we need to add both of the two last digits, and this should be the easiest example! All numbers that end in <u>00</u>, meaning they have a round number of hundreds, are divisible by 4! Therefore our option is: <em>9100.</em>

670_: In this case, we apply the same rule from the last example - and we choose <u>0</u>: <em>6700.</em>

(v) In order for a number to be divisible by 8, its last three digits need to be divisible by 8, or it can end in <u>000,</u> since every number having a round number of thousands is also divisible by 8.

1232_: 320 is divisible by 8 since 8 × 40 gives 320. Therefore, our smallest digit here is <u>0</u>: <em>12320.</em>

59_16: If we add a <u>0</u> here, the last three digits form the number 016, which is divisible by 8. Our choice is again zero: <em>59016.</em>

4642_: Unfortunately, 420 is not divisible by 8. The next number closest to 420 and which is divisible by 8 is 424. Its division by 8 gives 53. Therefore we should add <u>4</u>: <em>46424.</em>

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Answer:

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Step-by-step explanation:

Let the numbers be a and b.

<u>A positive number is 7 times another number:</u>

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<u>If 3 is added to both the numbers then one of the new number becomes 5 by 2 times the other new number:</u>

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<u>To solve this we substitute </u><u>a</u><u> with </u><u>7b</u><u> in the second equation:</u>

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<u>Then, finding a:</u>

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<u>So the numbers are</u> 7 and 1

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