# The four-digit number a55b is divisible by 6 without remainder. what is a and b?

Answers: 1

## Answers

The four-digit number a55b is divisible by 6 without remainder. what is a and b? ...

Answers: 1

The four-digit number a55b is divisible by 6 without remainder. what is a and b? ...

answer:

so many possible answer

step-by-step explanation:

if a= 3, or 6, or 9

b is 2

3552

6552

9552

if a is 1, or 4, or 7. then b is 4

1554

4554

7554

if a is 2, or 5, or 8. then b is 6

2556

5556

8556

if a is 3, or 6, or 9. then b is 4

3558

6558

9558

the rule is if the number is divisible by 2 or 3 it is divisible by 6

any even number is divisible by 2

add all the digit and if its sum must be divisible by 3.

ex. 9558

9+5+5+8= 27

27 is divisible by 3

not convinced?

then 2+7=9

9 is surely divisible by 3.

1001, 1221, 1331, 2816, 3465

Explanation:

if you want more just multiply 11 to a 3 digit number. ^_^

4444444444444444444444444

6000

6000÷2= 3000

6000÷3= 2000

6000÷4= 1500

6000÷6= 1000

6(first digit)-0(last digit)=6

9990

Step-by-step explanation:

9990/2=4995

9990/5=1998

999

Step-by-step explanation:

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answer:

3. 1000

4. 990

5. 60, 70, 80, 90

answer:

3.) 1, 000

Explanation: The common denominator that could be feasibly divided by the numbers 2, 5, and 10 is the number ended with 0. Since what has been asked here is the least 4-digit number then we could conclude that the digit that is belong with the thousands place is 1. And keep on filling the places of hundreds and tens with 0 to justify what was asked for, least 4-digit number.

FINAL ANSWER: 1, 0 0 0

4.) 990

Explanation: Same goes with the statement above. Since the common denominator that is likely to be divisible by 2, 5, and 10. We will fill the ones place with zero (0). In terms of the tens and hundreds places, considered what is being asked. It says that it should be the greatest 3-digit number, thus, think of the greatest number from the timeline (1-9). Therefore, you could conclude that the number 9 should be placed on the respective places of tens and hundreds.

FINAL ANSWER: 9 9 0

5.) 60, 70, 80, & 90

Explanation: The common denominator that are divisible by the numbers 2 and 5 do ended with zero (0). Since, the thing that was asked are numbers between 50 and 100 therefore:

60, 70, 80, and 90 are the numbers between 50 and 100 that are divisible by 2 and 5.

answer:

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Step-by-step explanation:

"Four digit numbers" does not allow leading zeros, so consider 1000 - 9999. If repetition is allowed, then every number ending with either 0 or 5 is divisible by five. There are 9000 values and 2 out of 10 (or 1 out of 5) of them are divisible by 5, so 9000/5 = 1800 are divisible by 5 -- if repetition is allowed.

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ANSWER:The first 4-digit number divisible by 11 is 1001.

EXPLANATION: This is sometimes also referred to as the smallest four digit number divisible by 11 or the lowest 4-digit number divisible by 11. What is the last four digit number divisible by 11? The last 4-digit number divisible by 11 is 9999.