0
It is assumed that the question requires each of the four digits to be used once and only once - ie permutations of {1, 2, 3, 4}.
Since 100 is divisible by 4, any number of hundreds is divisible by 4. So the test of divisibility by 4 reduces to whether the last two digits are divisible by 4.
For each pair of digits in the tens and units place that satisfy this requirement, the first two (thousands and hundreds) can be selected in two ways.
So there are two numbers ending in 12
two ending in 24
and two ending in 32
six in all.
What else can I help you with?
How many 9 digit numbers which are divisible by 4 can be formed by using the digits 1 to 5 if repetition of digits is not allowed?
None. Nine digit number cannot be formed using only five digits 1,2,3,4,5 in the case none of the digits can repeat. -------- Egad! I misread the question, apologies to the questioner. And thanks to Miroslav.
Can you make a prime number from the numbers 1 0 8?
A prime number is only divisible by 1 and itself. If all three digits have to be used then any number formed from them is divisible by 9 as the sum of the digits = 9. Thus no prime number can be formed. If only two of the three digits are used then it cannot end in 0 or 8 as it would be divisible by 2. This only leaves 81 where the digits again total 9 and thus is divisible by 9. None of the single digits 0,1,8 are prime. 8 is an even number. 1 is classed as unity, 0 is zero.
How many 5-digits-number are divisible by 4?
1
How many 5 digits can be formed with digits 012345 divisible by 3 with digits not repeated?
0+1+2+3+4+5 = 15 so, if a 5-digit number, formed from these digits, is divisible by 3 it must exclude 0 or 3. If it excludes 0, all 5*4*3*2*1 = 120 permutations are valid. If it excludes 3, then 4*4*3*2*1 = 96 permutations are valid (the others start with a zero and so are 4-digit numbers). That makes a total of 120+96 = 216 numbers.
Is 3 divisible by 518?
If sum of digits of a number is divisible by 3 then the number is divisible by 3. Sum of digits = 5+1+8 = 14 and 14 is not divisible by 3. So 518 is not divisible by 3.
How many 9 digit numbers which are divisible by 4 can be formed by using the digits 1 to 5 if repetition of digits is not allowed?
None. Nine digit number cannot be formed using only five digits 1,2,3,4,5 in the case none of the digits can repeat. -------- Egad! I misread the question, apologies to the questioner. And thanks to Miroslav.
Can you make a prime number from the numbers 1 0 8?
A prime number is only divisible by 1 and itself. If all three digits have to be used then any number formed from them is divisible by 9 as the sum of the digits = 9. Thus no prime number can be formed. If only two of the three digits are used then it cannot end in 0 or 8 as it would be divisible by 2. This only leaves 81 where the digits again total 9 and thus is divisible by 9. None of the single digits 0,1,8 are prime. 8 is an even number. 1 is classed as unity, 0 is zero.
Is 10000 divisible by three?
No. If a number is divisible by three, the sum of its digits will be divisible by three. Obviously, the sum of the digits of 10000 is 1, and 1 is not divisible by 3, so 10000 is not divisible by 3.
How many 5-digits-number are divisible by 4?
1
How many 5 digit no can b formed wit digits 1 2 3 4 5 6 which r divisible by 4 and digits not repeated?
Stop trying to cheat in math, it's your grade not mine...
How many 6-digit number can be formed from the digits 0 1 3 5 7and 9 which are divisible by 10 and no digit is repeated?
5*4*3*2*1 = 120
What are the divisibility rules for 144?
The digital root (sum of digit) must be divisible by 9, and the number formed by the last 4 digits must be divisible by 16. The second requirement ensures that the number is divisible by 16.
How many 5 digits can be formed with digits 012345 divisible by 3 with digits not repeated?
0+1+2+3+4+5 = 15 so, if a 5-digit number, formed from these digits, is divisible by 3 it must exclude 0 or 3. If it excludes 0, all 5*4*3*2*1 = 120 permutations are valid. If it excludes 3, then 4*4*3*2*1 = 96 permutations are valid (the others start with a zero and so are 4-digit numbers). That makes a total of 120+96 = 216 numbers.
Is 3 divisible by 518?
If sum of digits of a number is divisible by 3 then the number is divisible by 3. Sum of digits = 5+1+8 = 14 and 14 is not divisible by 3. So 518 is not divisible by 3.
How many 9 digit numbers which are divisible by 4 can be formed by using the digits 1 to 5 if repetition of digits is allowed?
Reqd no. is 9 -digits.. available digits are 1 2 3 4 5.. Numbers which are divisible by 4 can be determined by the last two digits.. from the given combinations,, the numbers div by 4 are 12, 24, 32, 44, 52 so.. the number's last two digits can be any of the above 5. Hence, we have to calculate the combinations for the first 7 digits.. Ans: (5C1)^7 * 5 ie 5^8 = 390625
How many 3 digit numbers can be formed from the digits 12345 if the digits cannot be repeated?
1 set
Is 1 trillion 17 a prime number?
Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.Since the sum of the digits is divisible by 3, the original number is also divisible by 3.
