The following bands/artists had at least three Christmas No 1s in the UK:
1998 Spice Girls - Goodbye
1997 Spice Girls - Too Much
1996 Spice Girls - 2 become 1
1967 The Beatles - Hello, Goodbye
1965 The Beatles - Day Tripper/We Can Work It Out
1964 The Beatles - I Feel Fine
1963 The Beatles - I Want to Hold Your Hand
(John Lennon and Paul McCartney also separately had Christmas No 1s, Lennon in 1980 with (Just Like) Starting Over and McCartney (with Wings) in 1978 with Mull of Kintyre/Girls' School)
1990 Cliff Richard - Saviour's Day
1988 Cliff Richard - Mistletoe and Wine
1960 Cliff Richard & The Shadows - I Love You
jive bunny
The Beatles three no 1s were in 1963 I want to hold your hand 1964 I feel fine 1965 daytripper/we can work it out.The Spice Girls were in 1996 2 become 1 1997 Too much and in 1998 Goodbye.
Do They Know It's Christmas by Live Aid
Three Wise Men
All I want for Christmas is my 2 front teeth. I Saw Three ships Come Sailing In We Three Kings
No. Any three consecutive numbers will have at least one of them which is divisible by 2, which means it cannot be prime. And since 1 is not considered a prime number, it cannot happen.
Yes, the product of any three consecutive numbers is divisible by 6. This is because among any three consecutive integers, at least one of them is even (ensuring divisibility by 2), and at least one of them is divisible by 3. Since 6 is the product of 2 and 3, the product of any three consecutive numbers is therefore divisible by 6.
That isn't possible. The three consecutive number are assumed to be integers; the sum of three consecutive integers is always a multiple of 3 (try it out).
Not sure what thress is. If three, then there is no answer since the sum (or product) of any three consecutive integers must be divisible by 3.
The smallest is 55.
That doesn't work. The number has to be divisible by three. Any three consecutive numbers add up to a multiple of three.
If the first number is 915, then the second will be 915 + 1 = 916 and the third number will be 915 + 2 = 917. Add 915, 916 and 917 to find the sum. If the question is, what are the three consecutive numbers whose sum is 915, then divide 915 by 3 to find the middle number which is 305. Since the consecutive numbers differ by 1, the numbers are 304, 305, and 306.
Since 57/3 = 19, we know that the middle number has to be 19. The other numbers will be the numbers contiguous to 19. Therefore, the three consecutive numbers which add up to 57 are: 18 + 19 + 20 = 57
Their sum is three times the middle number.
Any two consecutive numbers must comprise one odd and one even number, so their product must be even. Any three consecutive numbers must include two consecutive numbers so the result still applies.
The number of total outcomes on 3 tosses for a coin is 2 3, or 8. Since only 1 outcome is H, H, H, the probability of heads on three consecutive tosses of a coin is 1/8.
The sum of two odd numbers is even and the sum of an even number and an odd number is odd.So the sum of three consecutive odd numbers is odd but 270 is even.Ergo there are no three consecutive odd numbers that sum to 270.Three consecutive numbers that sum to 270 are 89, 90, 91.Three consecutive even numbers that sum to 270 are 88, 90, 92.